ANSWER: 4Y + 2X – 2

Home , Binomial (polynomial), Degree of a polynomial, Trinomial

Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER: 5-2 Dividing Polynomials

2 Simplify. 4. (2a – 4a – 8) ÷ (a + 1)

1. SOLUTION:

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) ANSWER: SOLUTION: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

ANSWER: 2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

eSolutions Manual - Powered by Cognero Page 1 ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

ANSWER: D.

SOLUTION: First rewrite the divisor so the x-term is first. Then 5 2 6. (y – 3y – 20) ÷ (y – 2) use long division.

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 2 8. (10x + 15x + 20) ÷ (5x + 5) SOLUTION: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

D. ANSWER: SOLUTION: First rewrite the divisor so the x-term is first. Then use long division. 2 9. (18a + 6a + 9) ÷ (3a – 2)

SOLUTION:

The correct choice is A. ANSWER: ANSWER: A

Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) 10. SOLUTION: SOLUTION:

ANSWER:

11. 2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: SOLUTION:

ANSWER: ANSWER: 3y + 5 Simplify

12. 10. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13.

SOLUTION: ANSWER:

11.

SOLUTION: ANSWER: x + 3y –2

14.

SOLUTION:

ANSWER: 3y + 5 ANSWER:

Simplify

12. 15. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13. ANSWER: 2 SOLUTION: 2a + b – 3

16.

SOLUTION:

ANSWER: x + 3y –2

14. ANSWER: SOLUTION: 4c2d2 – 6

17.

SOLUTION:

ANSWER:

15. ANSWER: SOLUTION: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: ANSWER: The average amount of energy saved per CFL bulb 2 2a + b – 3 is:

16.

SOLUTION: ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. ANSWER: SOLUTION: 2 2 4c d – 6 The average number of cookies produced per worker is:

17.

SOLUTION:

ANSWER:

Simplify. 2 ANSWER: 20. (a – 8a – 26) ÷ (a + 2) 3np – 6 + 7p SOLUTION: 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: ANSWER: The average amount of energy saved per CFL bulb is:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: ANSWER:

–b + 8

SOLUTION: The average number of cookies produced per worker is: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: Simplify. 3 2 2 z – 2z – 4 20. (a – 8a – 26) ÷ (a + 2) 5 3 2 SOLUTION: 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER: ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) 24. SOLUTION: SOLUTION:

ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION: 25. (g4 – 3g2 – 18) ÷ (g – 2)

SOLUTION:

ANSWER: z3 – 2z2 – 4

5 3 2 ANSWER: 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER: ANSWER:

27. 25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27.

SOLUTION: ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

28.

SOLUTION: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: 32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship Synthetic division: defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER:

6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) The depressed polynomial is . SOLUTION:

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an ANSWER: expression that represents the voltage.

SOLUTION:

ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these

instructions to a volunteer. The depressed polynomial is . • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the Since the volume of the box is the product of length, end? width and height, the width and length of box are (2x b. Explain the process you used to discover the + 3)(3x + 5). answer. SOLUTION: ANSWER: a. Let x be the number. 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If b. Sample answer: Let x be the number. Multiply the the current of the generator is I = t + 4, write an x by 4 to get 4x. Then add x +15 to the product to get expression that represents the voltage. 5x + 15. Divide the polynomial by x + 3. The quotient is 5. SOLUTION: ANSWER: a.5 b. Sample answer: Let x be the number. Synthetic division: Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is

the amount of money the company spent on advertising in hundreds of dollars and n is the number ANSWER: of subscriptions sold. 2 V(t) = t + 5t + 6 a. Perform the division indicated by .

34. ENTERTAINMENT A magician gives these b. About how many subscriptions will be sold if instructions to a volunteer. $1500 is spent on advertising? • Choose a number and multiply it by 4. SOLUTION: • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer.

SOLUTION: a. Let x be the number.

b. There are 15 hundreds in 1500. Substitute a = 15.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 b. Sample answer: Let x be the number. Therefore, about 2423 subscriptions will be sold. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + ANSWER:

3. The quotient is 5. a. 35. BUSINESS The number of magazine subscriptions b. about 2423 subscriptions

sold can be estimated by ,where a is Simplify. 4 4 the amount of money the company spent on 36. (x – y ) ÷ (x – y) advertising in hundreds of dollars and n is the number SOLUTION: of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: ANSWER: 2 2 a. (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

Therefore, about 2423 subscriptions will be sold. ANSWER: ANSWER:

a. b. about 2423 subscriptions 39. Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION: SOLUTION:

ANSWER: ANSWER: n2 – n – 1 2 2 (x + y )(x + y)

40. 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER:

ANSWER: 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

41.

SOLUTION:

ANSWER:

39.

SOLUTION: ANSWER: 4 3 2 3z – z + 2 z – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. ANSWER: b. SYMBOLIC Write an expression to represent n2 – n – 1 the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete 40. model check with your algebraic model? SOLUTION: SOLUTION: a.

ANSWER:

41. The width is x + 3.

SOLUTION: b.

ANSWER: 4 3 2 Yes, the concrete model checks with the algebraic 3z – z + 2 z – 4z + 9– model.

ANSWER: 42. MULTIPLE REPRESENTATIONS Consider a 2 a. rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes The width is x + 3. 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 b. 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your c. reasoning. SOLUTION: Use synthetic division.

Yes, the concrete model checks with the algebraic model.

ANSWER: The remainder is 26. So, Sharon is correct. Jamal a. actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION:

If a polynomial divided by a binomial has no The width is x + 3. remainder, then the polynomial has two factors: the binomial and the quotient. 2 b. 2x + 7x + 3 ÷ (2x + 1) ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems c. in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION:

yes Sample answer: For example:

In this exercise, the degree of the dividend is 2. The The remainder is 26. So, Sharon is correct. Jamal degree of the divisor and the quotient are each 1. actually divided by x + 3. The degree of the quotient plus the degree of the divisor equals the degree of the dividend. ANSWER: Sample answer: Sharon; Jamal actually divided by x ANSWER: + 3. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the 44. CHALLENGE If a polynomial is divided by a dividend. binomial and the remainder is 0, what does this tell you about the relationship between the binomial and 46. OPEN ENDED Write a quotient of two polynomials the polynomial? for which the remainder is 3. SOLUTION: SOLUTION: If a polynomial divided by a binomial has no Sample answer: Begin by multiplying two binomials remainder, then the polynomial has two factors: the such as (x + 2)(x + 3) which simplifies to x2 + 5x +

binomial and the quotient. 6. In order to get a remainder of 3 when divided, add 2 ANSWER: 3 to the trinomial to get x + 5x + 9. When divided, The binomial is a factor of the polynomial. there will be a remainder of 3. The quotient of two

45. REASONING Review any of the division problems polynomials is . in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? ANSWER: SOLUTION: Sample answer: Sample answer: For example:

47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

SOLUTION:

ANSWER: ANSWER: Sample answer: The degree of the quotient plus the does not belong with the other three. The other degree of the divisor equals the degree of the dividend. three expressions are polynomials. Since the

46. OPEN ENDED Write a quotient of two polynomials denominator of contains a variable, it is not a for which the remainder is 3. polynomial. SOLUTION: Sample answer: Begin by multiplying two binomials 48. WRITING IN MATH Use the information at the 2 beginning of the lesson to write assembly instruction such as (x + 2)(x + 3) which simplifies to x + 5x + using the division of polynomials to make a paper 6. In order to get a remainder of 3 when divided, add cover for your textbook. 2 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two SOLUTION: Sample answer: By dividing the area of the paper polynomials is . 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book ANSWER: jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting Sample answer: 14x, we are left with 6 inches. Half of this length is the width of each flap.

47. CCSS ARGUMENTS Identify the expression that ANSWER: does not belong with the other three. Explain your 2 reasoning. Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women? SOLUTION:

A does not belong with the other three. The other

three expressions are polynomials. Since the B denominator of contains a variable, it is not a C polynomial.

D ANSWER:

48. WRITING IN MATH Use the information at the The correct choice is A. beginning of the lesson to write assembly instruction using the division of polynomials to make a paper ANSWER: cover for your textbook. A SOLUTION: 50. SAT/ACT Which polynomial has degree 3? Sample answer: By dividing the area of the paper 3 2 4 2 A x + x –2x 140x + 60x by the height of the book jacket 10x, the 2 quotient of 14x + 6 provides the length of the book B –2x – 3x + 4 jacket. The front and back cover are each 6x units C 3x – 3 long and the spine is 2x units long. Then, subtracting D x2 + x + 123 3 14x, we are left with 6 inches. Half of this length is E 1 + x + x the width of each flap. SOLUTION: ANSWER: To determine the degree of a polynomial, look at the 2 Sample answer: By dividing 140x + 60x by 10x, the term with the greatest exponent. quotient of 14x + 6 provides the length of the book The correct choice is E. jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. ANSWER: E 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number 51. GRIDDED RESPONSE In the figure below, of women? m + n + p = ?

C SOLUTION: D Sum of the exterior angles of a triangle is . So, SOLUTION: Total number of employees: x + 3 ANSWER: Number of women: x 360 Ratio of the total number of employees to the number of women is: 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x H 2 The correct choice is A. –10x + 17x 2 G –10x ANSWER: 2 J 2x + 17x A SOLUTION: 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 2 B –2x – 3x + 4 C 3x – 3 2 3 The correct choice is H. D x + x + 12 3 E 1 + x + x ANSWER: H SOLUTION: To determine the degree of a polynomial, look at the Simplify. 3 2 3 term with the greatest exponent. 53. (5x + 2x – 3x + 4) – (2x – 4x) The correct choice is E. SOLUTION: ANSWER: Use the Distributive Property and then combine like E terms.

51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 3 2 3x + 2x + x + 4 SOLUTION: Sum of the exterior angles of a triangle is . 3 2 54. (2y – 3y + 8) + (3y – 6y) So, SOLUTION: ANSWER: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = ANSWER: 2 F 2x 2y 3 + 3y 2 – 9y + 8 H –10x2 + 17x 2 55. 4a(2a – 3) + 3a(5a – 4) G –10x 2 SOLUTION: J 2x + 17x Use the Distributive Property and then combine like SOLUTION: terms.

ANSWER: 2 The correct choice is H. 23a – 24a

ANSWER: 56. (c + d)(c – d)(2c – 3d) H SOLUTION: Simplify. Use the FOIL method and then combine like terms. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: Use the Distributive Property and then combine like ANSWER: terms. 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 3x3 + 2x2 + x + 4 ANSWER: 3 2 54. (2y – 3y + 8) + (3y – 6y) 8x5y 8z 3

SOLUTION: 2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: Use the Distributive Property and then combine like ANSWER: terms.

ANSWER: 59. LANDSCAPING Amado wants to plant a garden 2 and surround it with decorative stones. He has 23a – 24a enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to 56. (c + d)(c – d)(2c – 3d) cover no more than 240 square feet. What could the SOLUTION: width of his garden be? Use the FOIL method and then combine like terms. SOLUTION: Let x be the length and y be the width of the garden.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the expression. The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is ANSWER: 5 8 3 8x y z ANSWER: 2 –2 2 2 0 to 10 ft or 24 to 34 ft 58. (3ab ) (2a b) SOLUTION: Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

cover no more than 240 square feet. What could the –3 ± width of his garden be?

SOLUTION: 2 61. x – 8x – 3 = 0 Let x be the length and y be the width of the garden. SOLUTION:

The solution of the inequality is ANSWER: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is 62. 2x2 + 6x + 5 = 0

SOLUTION: ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located. ANSWER:

–3 ± 63. SOLUTION: 2 61. x – 8x – 3 = 0 The sign of f (x) changes between the x values –6 SOLUTION: and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

ANSWER: 64. SOLUTION: The sign of f (x) changes between the x values 62. 2x2 + 6x + 5 = 0 between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and SOLUTION: between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3. ANSWER: ANSWER:

between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 State the consecutive integers between which minutes and perform a small landscape job in 90 the zeros of each quadratic function are located. minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per 63. landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed SOLUTION: and completed landscape jobs per week that will The sign of f (x) changes between the x values –6 maximize income. Then find the maximum income. and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and SOLUTION: between –3 and –2. Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most ANSWER: 3 landscaping jobs per day, L ≤ 15. Since he works between –6 and –5; between –3 and –2 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100.

64. If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total SOLUTION: of 10 hours per day. So, for the week, if he does 15 The sign of f (x) changes between the x values landscaping jobs, he can mow 55 lawns. between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and To maximize the earnings, write a function that between 2 and 3. relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and ANSWER: $125 for each small landscaping job, F(M, L) = 35M between 0 and 1; between 2 and 3 + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

65. (M, L) F(M, L) = 35M + F(M, L) SOLUTION: 125L The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the (100, 0) F(M, L) = 35(100) + 3500 zeros of the quadratic function lies between –1 and 0, 125(0) and between 2 and 3. (55, 15) F(M, L) = 35(55) + 125 3800 ANSWER: (15) between –1 and 0; between 2 and 3 (0, 15) F(M, L) = 35(0) + 125 1875 66. BUSINESS A landscaper can mow a lawn in 30 (15) minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per 15 landscape jobs and 55 lawns; $3800 landscape job. He cannot do more than 3 landscape ANSWER: jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will 15 landscape jobs and 55 lawns; $3800 maximize income. Then find the maximum income. Find each value if f (x) = 4x + 3, g(x) = –x2, and SOLUTION: h(x) = –2x2 – 2x + 4. Let M be the lawns mowed and L be the small 67. f (–6) landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works SOLUTION: 10 hours per day, 5 days a week, he can do at most Substitute –6 for x in f (x). 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. ANSWER: –21 To maximize the earnings, write a function that relates the number and charge for each type of job. 68. g(–8) Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M SOLUTION: + 125L. Next, substitute in the key values of M and Substitute –8 for x in g(x). L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) 125L ANSWER: –64 (100, 0) F(M, L) = 35(100) + 3500 125(0) 69. h(3) SOLUTION: (55, 15) F(M, L) = 35(55) + 125 3800 Substitute 3 for x in h(x). (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800

ANSWER: ANSWER: 15 landscape jobs and 55 lawns; $3800 –20

Find each value if f (x) = 4x + 3, g(x) = –x2, and 70. f (c) 2 h(x) = –2x – 2x + 4. SOLUTION: 67. f (–6) Substitute c for x in f (x). SOLUTION: Substitute –6 for x in f (x).

ANSWER: 4c + 3 71. g(3d) ANSWER: –21 SOLUTION: Substitute 3d for x in g(x).

68. g(–8)

SOLUTION: Substitute –8 for x in g(x). ANSWER: 2 –9d

72. h(2b + 1) ANSWER: –64 SOLUTION: Substitute 2b + 1 for x in h(x). 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –8b2 – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

Simplify.

SOLUTION:

ANSWER: ANSWER:

4y + 2x – 2

2 2 –1 2 2. (3a b – 6ab + 5ab )(ab) 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3)

SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER: ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) 4 3 2 5. (3z – 6z – 9z + 3z – 6) ÷ (z + 3) SOLUTION: SOLUTION:

ANSWER: ANSWER: 5-2 Dividing Polynomials y 4 + 2y3 + 4y2 + 5y + 10

5 2 6. (y – 3y – 20) ÷ (y – 2) 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? SOLUTION:

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

MULTIPLE CHOICE 7. Which expression is equal The correct choice is A. 2 –1 to (x + 3x – 9)(4 – x) ? ANSWER: A. A B . Simplify.

C. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: D.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

eSolutions Manual - Powered by Cognero Page 2

2 The correct choice is A. 9. (18a + 6a + 9) ÷ (3a – 2) ANSWER: SOLUTION: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

10.

SOLUTION: ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

11.

ANSWER: SOLUTION:

10.

SOLUTION:

ANSWER: 3y + 5 Simplify

12.

ANSWER: SOLUTION:

ANSWER: 11. 3a2b – 2ab2 SOLUTION:

13.

SOLUTION:

ANSWER: ANSWER: 3y + 5 x + 3y –2 Simplify 14. 12. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2 ANSWER:

13.

SOLUTION: 15.

SOLUTION:

ANSWER: x + 3y –2

ANSWER: 14. 2 2a + b – 3

SOLUTION: 16.

SOLUTION:

ANSWER:

15. ANSWER: 4c2d2 – 6 SOLUTION:

17.

SOLUTION:

ANSWER: 2 2a + b – 3 ANSWER: 3np – 6 + 7p 16. 18. ENERGY Compact fluorescent light (CFL) bulbs SOLUTION: reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is: ANSWER: 4c2d2 – 6

17. ANSWER: –b + 8 SOLUTION: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: ANSWER: The average number of cookies produced per worker 3np – 6 + 7p is:

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of ANSWER: energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by ANSWER: w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker 3 2 is: 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

ANSWER: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2)

SOLUTION: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1)

SOLUTION:

ANSWER: ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) 21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: SOLUTION:

ANSWER: ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) 24. SOLUTION: SOLUTION:

ANSWER: 3 2 z – 2z – 4 ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: 25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

24.

SOLUTION: 2 26. (6a – 3a + 9) ÷ (3a – 2)

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

27.

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

SOLUTION:

ANSWER:

27.

SOLUTION: 28.

SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: ANSWER:

28. Synthetic division: SOLUTION:

ANSWER:

ANSWER: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: Synthetic division:

Synthetic division:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION: ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the SOLUTION: box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

The depressed polynomial is .

ANSWER: Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three ANSWER: dimensions always have a particular relationship 2x + 3, 3x + 5 defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is 33. PHYSICS The voltage V is related to current I and always x + 2. What are the width and length of the power P by the equation . The power of a box? 3 2 SOLUTION: generator is modeled by P(t) = t + 9t + 26t + 24. If the current of the generator is I = t + 4, write an Divide the function by the height (x + 2) to find the expression that represents the voltage. length and width. Use synthetic division. SOLUTION:

Synthetic division:

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5). ANSWER: 2 V(t) = t + 5t + 6 ANSWER: ENTERTAINMENT 2x + 3, 3x + 5 34. A magician gives these instructions to a volunteer. 33. PHYSICS The voltage V is related to current I and • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the power P by the equation . The power of a product you found. • Now divide by the sum of your number and 3. 3 2 generator is modeled by P(t) = t + 9t + 26t + 24. If a. What number will the volunteer always have at the the current of the generator is I = t + 4, write an end? expression that represents the voltage. b. Explain the process you used to discover the

SOLUTION: answer. SOLUTION: a. Let x be the number.

Synthetic division:

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: ANSWER: 2 V(t) = t + 5t + 6 a.5 b. Sample answer: Let x be the number. 34. ENTERTAINMENT A magician gives these Multiply the x by 4 to get 4x. Then add x + 15 to the instructions to a volunteer. product to get 5x + 15. Divide the polynomial by x + • Choose a number and multiply it by 4. 3. The quotient is 5. • Then add the sum of your number and 15 to the product you found. 35. BUSINESS The number of magazine subscriptions • Now divide by the sum of your number and 3. sold can be estimated by ,where a is a. What number will the volunteer always have at the end? the amount of money the company spent on b. Explain the process you used to discover the advertising in hundreds of dollars and n is the number answer. of subscriptions sold.

SOLUTION: a. Perform the division indicated by . a. Let x be the number. b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION:

a. b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the b. There are 15 hundreds in 1500. Substitute a = 15. product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by . Therefore, about 2423 subscriptions will be sold. b. About how many subscriptions will be sold if ANSWER: $1500 is spent on advertising?

SOLUTION: a. b. about 2423 subscriptions a. Simplify. 36. (x4 – y4) ÷ (x – y)

SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15.

ANSWER: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER: ANSWER:

a. b. about 2423 subscriptions 3 2 2 –1 38. (a b – a b + 2b)(–ab) Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION: SOLUTION:

ANSWER:

ANSWER: 2 2 (x + y )(x + y) 39.

3 2 2 SOLUTION: 37. (28c d – 21cd ) ÷ (14cd)

SOLUTION:

ANSWER: ANSWER: n2 – n – 1

3 2 2 –1 38. (a b – a b + 2b)(–ab) 40. SOLUTION: SOLUTION:

ANSWER:

39.

SOLUTION:

41.

SOLUTION:

ANSWER:

n2 – n – 1

40. ANSWER: 4 3 2 SOLUTION: 3z – z + 2 z – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. ANSWER: c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION:

41. a.

SOLUTION:

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9– The width is x + 3.

42. MULTIPLE REPRESENTATIONS Consider a b. 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. c. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: Yes, the concrete model checks with the algebraic model. a. ANSWER: a.

The width is x + 3.

b. The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1) c.

Yes, the concrete model checks with the algebraic yes model. 43. ERROR ANALYSIS Sharon and Jamal are dividing ANSWER: 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the a. remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal The width is x + 3. actually divided by x + 3.

ANSWER: b. 2x 2 + 7x + 3 ÷ (2x + 1) Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a c. binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? yes SOLUTION: 43. ERROR ANALYSIS Sharon and Jamal are dividing If a polynomial divided by a binomial has no 3 2 remainder, then the polynomial has two factors: the 2x – 4x + 3x – 1 by x – 3. Sharon claims that the binomial and the quotient. remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your ANSWER: reasoning. The binomial is a factor of the polynomial. SOLUTION: Use synthetic division. 45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient?

SOLUTION: Sample answer: For example:

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION: In this exercise, the degree of the dividend is 2. The If a polynomial divided by a binomial has no degree of the divisor and the quotient are each 1. remainder, then the polynomial has two factors: the The degree of the quotient plus the degree of the binomial and the quotient. divisor equals the degree of the dividend. ANSWER: ANSWER: The binomial is a factor of the polynomial. Sample answer: The degree of the quotient plus the 45. REASONING Review any of the division problems degree of the divisor equals the degree of the

in this lesson. What is the relationship between the dividend. degrees of the dividend, the divisor, and the quotient? 46. OPEN ENDED Write a quotient of two polynomials SOLUTION: for which the remainder is 3. Sample answer: For example: SOLUTION:

Sample answer: Begin by multiplying two binomials

such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add 2 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two

polynomials is .

ANSWER:

Sample answer:

In this exercise, the degree of the dividend is 2. The 47. CCSS ARGUMENTS Identify the expression that degree of the divisor and the quotient are each 1. does not belong with the other three. Explain your The degree of the quotient plus the degree of the reasoning. divisor equals the degree of the dividend.

ANSWER: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend. SOLUTION: 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. does not belong with the other three. The other SOLUTION: three expressions are polynomials. Since the Sample answer: Begin by multiplying two binomials denominator of contains a variable, it is not a such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add polynomial. 2 3 to the trinomial to get x + 5x + 9. When divided, ANSWER: there will be a remainder of 3. The quotient of two does not belong with the other three. The other polynomials is . three expressions are polynomials. Since the ANSWER: denominator of contains a variable, it is not a

Sample answer: polynomial.

48. WRITING IN MATH Use the information at the 47. CCSS ARGUMENTS Identify the expression that beginning of the lesson to write assembly instruction does not belong with the other three. Explain your using the division of polynomials to make a paper reasoning. cover for your textbook. SOLUTION: Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units SOLUTION: long and the spine is 2x units long. Then, subtracting does not belong with the other three. The other 14x, we are left with 6 inches. Half of this length is the width of each flap. three expressions are polynomials. Since the ANSWER: denominator of contains a variable, it is not a 2 Sample answer: By dividing 140x + 60x by 10x, the polynomial. quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 ANSWER: inches. Half of this length is the width of each flap.

does not belong with the other three. The other 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number three expressions are polynomials. Since the of women?

denominator of contains a variable, it is not a A polynomial. B 48. WRITING IN MATH Use the information at the

beginning of the lesson to write assembly instruction C using the division of polynomials to make a paper cover for your textbook. D SOLUTION: Sample answer: By dividing the area of the paper SOLUTION: 140x2 + 60x by the height of the book jacket 10x, the Total number of employees: x + 3 quotient of 14x + 6 provides the length of the book Number of women: x jacket. The front and back cover are each 6x units Ratio of the total number of employees to the number long and the spine is 2x units long. Then, subtracting of women is: 14x, we are left with 6 inches. Half of this length is the width of each flap. The correct choice is A. ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the ANSWER: quotient of 14x + 6 provides the length of the book A jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 49. An office employs x women and 3 men. What is the 2 ratio of the total number of employees to the number B –2x – 3x + 4 of women? C 3x – 3 D x2 + x + 123 A 3 E 1 + x + x

B SOLUTION: To determine the degree of a polynomial, look at the

C term with the greatest exponent. The correct choice is E.

D ANSWER: SOLUTION: E Total number of employees: x + 3 51. GRIDDED RESPONSE In the figure below, Number of women: x m + n + p = ? Ratio of the total number of employees to the number of women is:

The correct choice is A.

ANSWER: SOLUTION: A Sum of the exterior angles of a triangle is . 50. SAT/ACT Which polynomial has degree 3? So, A x3 + x2 –2x4 ANSWER: 2 B –2x – 3x + 4 360 C 3x – 3 2 3 D x + x + 12 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 3 2 E 1 + x + x F 2x 2 SOLUTION: H –10x + 17x 2 To determine the degree of a polynomial, look at the G –10x term with the greatest exponent. J 2x2 + 17x The correct choice is E. SOLUTION: ANSWER: E

51. GRIDDED RESPONSE In the figure below, m + n + p = ? The correct choice is H.

ANSWER: H Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: SOLUTION: Sum of the exterior angles of a triangle is . Use the Distributive Property and then combine like

So, terms. ANSWER:

360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x H 2 –10x + 17x ANSWER: 2 G –10x 3x3 + 2x2 + x + 4 J 2x2 + 17x

3 2 SOLUTION: 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

The correct choice is H. ANSWER: ANSWER: 2y 3 + 3y 2 – 9y + 8 H 55. 4a(2a – 3) + 3a(5a – 4) Simplify. 3 2 3 SOLUTION: 53. (5x + 2x – 3x + 4) – (2x – 4x) Use the Distributive Property and then combine like SOLUTION: terms. Use the Distributive Property and then combine like terms.

ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: ANSWER: Use the FOIL method and then combine like terms. 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) ANSWER: 3 2 2 3 SOLUTION: 2c – 3c d – 2cd + 3d 2 2 3 57. (xy) (2xy z) SOLUTION: ANSWER: Apply the properties of exponents to simplify the 2y 3 + 3y 2 – 9y + 8 expression.

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: Use the Distributive Property and then combine like ANSWER: 5 8 3 terms. 8x y z

2 –2 2 2 58. (3ab ) (2a b) ANSWER: SOLUTION: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms.

ANSWER:

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3 LANDSCAPING 2 2 3 59. Amado wants to plant a garden 57. (xy) (2xy z) and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a SOLUTION: perimeter of 68 feet, but he wants the garden to Apply the properties of exponents to simplify the cover no more than 240 square feet. What could the expression. width of his garden be?

SOLUTION: Let x be the length and y be the width of the garden. ANSWER:

8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is ANSWER:

ANSWER: 0 to 10 ft or 24 to 34 ft 59. LANDSCAPING Amado wants to plant a garden Solve each equation by completing the square. and surround it with decorative stones. He has 60. x2 + 6x + 2 = 0 enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to SOLUTION: cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

The solution of the inequality is –3 ± . Since y represent the 2 width of the garden and the sum of the length and 61. x – 8x – 3 = 0 width is 34, the width of the garden is SOLUTION:

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION: ANSWER:

62. 2x2 + 6x + 5 = 0 SOLUTION:

ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 SOLUTION: ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

ANSWER: 63. SOLUTION: The sign of f (x) changes between the x values –6 2 62. 2x + 6x + 5 = 0 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and SOLUTION: between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3. ANSWER: ANSWER: between 0 and 1; between 2 and 3

State the consecutive integers between which the zeros of each quadratic function are located. 65. 63. SOLUTION: The sign of f (x) changes between the x values SOLUTION: between –1 and 0, and between 2 and 3. So the The sign of f (x) changes between the x values –6 zeros of the quadratic function lies between –1 and 0, and –5, and between –3 and –2. So the zeros of the and between 2 and 3. quadratic function lies between –6 and –5, and between –3 and –2. ANSWER: between –1 and 0; between 2 and 3 ANSWER: between –6 and –5; between –3 and –2 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per 64. landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed SOLUTION: and completed landscape jobs per week that will The sign of f (x) changes between the x values maximize income. Then find the maximum income. between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and SOLUTION: between 2 and 3. Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most ANSWER: 3 landscaping jobs per day, L ≤ 15. Since he works between 0 and 1; between 2 and 3 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100.

65. If the landscaper does 3 landscaping jobs per day, he SOLUTION: can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 The sign of f (x) changes between the x values landscaping jobs, he can mow 55 lawns. between –1 and 0, and between 2 and 3. So the

zeros of the quadratic function lies between –1 and 0, To maximize the earnings, write a function that and between 2 and 3. relates the number and charge for each type of job. ANSWER: Since he charges $35 for each lawn mowed and between –1 and 0; between 2 and 3 $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and 66. BUSINESS A landscaper can mow a lawn in 30 L to determine the earnings for each combination of minutes and perform a small landscape job in 90 jobs. minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per (M, L) F(M, L) = 35M + F(M, L) landscape job. He cannot do more than 3 landscape 125L jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will (100, 0) F(M, L) = 35(100) + 3500 maximize income. Then find the maximum income. 125(0) SOLUTION: (55, 15) F(M, L) = 35(55) + 125 3800 Let M be the lawns mowed and L be the small (15) landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L 15. Since he works ≤ (0, 15) F(M, L) = 35(0) + 125 1875 10 hours per day, 5 days a week, he can do at most (15) 20 mowing jobs per day or 100 per week. So, M ≤ 100. 15 landscape jobs and 55 lawns; $3800 If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total ANSWER: of 10 hours per day. So, for the week, if he does 15 15 landscape jobs and 55 lawns; $3800 landscaping jobs, he can mow 55 lawns. Find each value if f (x) = 4x + 3, g(x) = –x2, and To maximize the earnings, write a function that h(x) = –2x2 – 2x + 4. relates the number and charge for each type of job. 67. f (–6) Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M SOLUTION: + 125L. Next, substitute in the key values of M and Substitute –6 for x in f (x). L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) 125L ANSWER: (100, 0) F(M, L) = 35(100) + 3500 –21 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 68. g(–8) (15) SOLUTION: Substitute –8 for x in g(x). (0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800 ANSWER: ANSWER: –64 15 landscape jobs and 55 lawns; $3800 69. h(3) Find each value if f (x) = 4x + 3, g(x) = –x2, and SOLUTION: h(x) = –2x2 – 2x + 4. 67. f (–6) Substitute 3 for x in h(x).

SOLUTION: Substitute –6 for x in f (x).

ANSWER: ANSWER: –20 –21 70. f (c) SOLUTION: 68. g(–8) Substitute c for x in f (x). SOLUTION: Substitute –8 for x in g(x).

ANSWER: 4c + 3 ANSWER: 71. g(3d) –64 SOLUTION: 69. h(3) Substitute 3d for x in g(x). SOLUTION: Substitute 3 for x in h(x).

ANSWER: 2 –9d

72. h(2b + 1) ANSWER: SOLUTION: –20 Substitute 2b + 1 for x in h(x).

70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: ANSWER: –8b2 – 12b 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 Simplify. 4. (2a – 4a – 8) ÷ (a + 1)

1. SOLUTION:

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) ANSWER: SOLUTION: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

ANSWER: 2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: 5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

ANSWER: D.

SOLUTION: First rewrite the divisor so the x-term is first. Then 5 2 use long division. 6. (y – 3y – 20) ÷ (y – 2)

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 2 8. (10x + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

D. ANSWER:

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division. 2 9. (18a + 6a + 9) ÷ (3a – 2)

SOLUTION:

The correct choice is A. ANSWER: 5-2 DividingANSWER: Polynomials A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) 10.

SOLUTION: SOLUTION:

ANSWER:

11.

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: SOLUTION:

ANSWER: ANSWER: 3y + 5

Simplify

12.

10. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13. eSolutions Manual - Powered by Cognero Page 3 SOLUTION:

ANSWER:

11.

ANSWER: SOLUTION: x + 3y –2

14.

SOLUTION:

ANSWER: 3y + 5 ANSWER:

Simplify

12. 15. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13. ANSWER: 2 SOLUTION: 2a + b – 3

16.

SOLUTION:

ANSWER: x + 3y –2

14. ANSWER: SOLUTION: 4c2d2 – 6

17.

SOLUTION:

ANSWER:

15. ANSWER: SOLUTION: 3np – 6 + 7p 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: ANSWER: The average amount of energy saved per CFL bulb 2 2a + b – 3 is:

16.

SOLUTION: ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. ANSWER: SOLUTION: 4c2d2 – 6 The average number of cookies produced per worker is:

17.

SOLUTION:

ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: ANSWER: –b + 8

SOLUTION: The average number of cookies produced per worker 4 3 2 –1 is: 22. (z – 3z + 2z – 4z + 4)(z – 1)

SOLUTION:

ANSWER:

ANSWER: Simplify. z3 – 2z2 – 4 2 20. (a – 8a – 26) ÷ (a + 2) 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER: ANSWER:

3 2 21. (b – 4b + b – 2) ÷ (b + 1) 24. SOLUTION: SOLUTION:

ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1)

SOLUTION: 25. (g4 – 3g2 – 18) ÷ (g – 2)

SOLUTION:

ANSWER: z3 – 2z2 – 4 ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER: ANSWER:

27. 25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27.

SOLUTION: ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

28.

6 4 2 –1 SOLUTION: 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: 32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can Synthetic division: 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 The depressed polynomial is .

SOLUTION:

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an ANSWER: expression that represents the voltage. SOLUTION:

32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship Synthetic division: defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer. The depressed polynomial is . • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? Since the volume of the box is the product of length, b. Explain the process you used to discover the width and height, the width and length of box are (2x answer. + 3)(3x + 5). SOLUTION: a. Let x be the number. ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a b. Sample answer: Let x be the number. Multiply the generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If x by 4 to get 4x. Then add x +15 to the product to get the current of the generator is I = t + 4, write an 5x + 15. Divide the polynomial by x + 3. The quotient expression that represents the voltage. is 5. SOLUTION: ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the Synthetic division: product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is

the amount of money the company spent on

advertising in hundreds of dollars and n is the number of subscriptions sold. ANSWER: 2 V(t) = t + 5t + 6 a. Perform the division indicated by . b. About how many subscriptions will be sold if 34. ENTERTAINMENT A magician gives these $1500 is spent on advertising? instructions to a volunteer. • Choose a number and multiply it by 4. SOLUTION: • Then add the sum of your number and 15 to the product you found. a. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b . Explain the process you used to discover the answer.

SOLUTION: a. Let x be the number. b. There are 15 hundreds in 1500. Substitute a = 15.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER:

a.5 Therefore, about 2423 subscriptions will be sold. b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the ANSWER: product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. a.

35. BUSINESS The number of magazine subscriptions b. about 2423 subscriptions

sold can be estimated by ,where a is Simplify. 36. (x4 – y4) ÷ (x – y) the amount of money the company spent on advertising in hundreds of dollars and n is the number SOLUTION: of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising?

SOLUTION: ANSWER: 2 2 (x + y )(x + y) a.

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

Therefore, about 2423 subscriptions will be sold. ANSWER: ANSWER:

a. b. about 2423 subscriptions 39. Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION:

SOLUTION:

ANSWER: ANSWER: n2 – n – 1 2 2 (x + y )(x + y)

40. 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

41.

SOLUTION:

ANSWER:

39. ANSWER: SOLUTION:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. ANSWER: b. SYMBOLIC Write an expression to represent 2 the model. n – n – 1 c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete 40. model check with your algebraic model? SOLUTION: SOLUTION: a.

ANSWER:

The width is x + 3. 41.

SOLUTION: b.

ANSWER: Yes, the concrete model checks with the algebraic 3z 4 – z 3 + 2 z2 – 4z + 9– model. ANSWER: 42. MULTIPLE REPRESENTATIONS Consider a a. 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

Yes, the concrete model checks with the algebraic model.

ANSWER: The remainder is 26. So, Sharon is correct. Jamal a. actually divided by x + 3. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the The width is x + 3. binomial and the quotient.

b. 2x 2 + 7x + 3 ÷ (2x + 1) ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems c. in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: yes Sample answer: For example:

polynomials is . 45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? ANSWER:

SOLUTION: Sample answer: Sample answer: For example: 47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. denominator of contains a variable, it is not a The degree of the quotient plus the degree of the polynomial. divisor equals the degree of the dividend. ANSWER: ANSWER: Sample answer: The degree of the quotient plus the does not belong with the other three. The other degree of the divisor equals the degree of the dividend. three expressions are polynomials. Since the denominator of contains a variable, it is not a 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. polynomial. SOLUTION: 48. WRITING IN MATH Use the information at the Sample answer: Begin by multiplying two binomials beginning of the lesson to write assembly instruction such as (x + 2)(x + 3) which simplifies to x2 + 5x + using the division of polynomials to make a paper 6. In order to get a remainder of 3 when divided, add cover for your textbook. 2 3 to the trinomial to get x + 5x + 9. When divided, SOLUTION: there will be a remainder of 3. The quotient of two Sample answer: By dividing the area of the paper 2 polynomials is . 140x + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units ANSWER: long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is Sample answer: the width of each flap.

47. CCSS ARGUMENTS Identify the expression that ANSWER: 2 does not belong with the other three. Explain your Sample answer: By dividing 140x + 60x by 10x, the reasoning. quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6

inches. Half of this length is the width of each flap. 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women? SOLUTION: A does not belong with the other three. The other

B three expressions are polynomials. Since the

denominator of contains a variable, it is not a C polynomial. D ANSWER: SOLUTION: does not belong with the other three. The other Total number of employees: x + 3 three expressions are polynomials. Since the Number of women: x Ratio of the total number of employees to the number denominator of contains a variable, it is not a of women is: polynomial.

48. WRITING IN MATH Use the information at the The correct choice is A. beginning of the lesson to write assembly instruction using the division of polynomials to make a paper ANSWER: cover for your textbook. A SOLUTION: 50. SAT/ACT Which polynomial has degree 3? Sample answer: By dividing the area of the paper A x3 + x2 –2x4 2 2 140x + 60x by the height of the book jacket 10x, the B –2x – 3x + 4 quotient of 14x + 6 provides the length of the book C 3x – 3 jacket. The front and back cover are each 6x units 2 3 long and the spine is 2x units long. Then, subtracting D x + x + 12 3 14x, we are left with 6 inches. Half of this length is E 1 + x + x the width of each flap. SOLUTION: ANSWER: To determine the degree of a polynomial, look at the 2 Sample answer: By dividing 140x + 60x by 10x, the term with the greatest exponent. quotient of 14x + 6 provides the length of the book The correct choice is E. jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. ANSWER: E 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number 51. GRIDDED RESPONSE In the figure below, of women? m + n + p = ?

C SOLUTION:

D Sum of the exterior angles of a triangle is . So, SOLUTION: Total number of employees: x + 3 ANSWER: Number of women: x 360 Ratio of the total number of employees to the number 2 2 of women is: 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2 F 2x H –10x2 + 17x The correct choice is A. 2 G –10x ANSWER: J 2x2 + 17x A SOLUTION: 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 B 2 –2x – 3x + 4 C 3x – 3 The correct choice is H. D x2 + x + 123 3 E 1 + x + x ANSWER: H SOLUTION: To determine the degree of a polynomial, look at the Simplify. 3 2 3 term with the greatest exponent. 53. (5x + 2x – 3x + 4) – (2x – 4x) The correct choice is E. SOLUTION: ANSWER: Use the Distributive Property and then combine like E terms.

51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 3x3 + 2x2 + x + 4 SOLUTION: 3 2 Sum of the exterior angles of a triangle is . 54. (2y – 3y + 8) + (3y – 6y) So, SOLUTION: ANSWER: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = ANSWER: 2 3 2 F 2x 2y + 3y – 9y + 8 2 H –10x + 17x 55. 4a(2a – 3) + 3a(5a – 4) 2 G –10x SOLUTION: 2 J 2x + 17x Use the Distributive Property and then combine like SOLUTION: terms.

ANSWER:

2 The correct choice is H. 23a – 24a

ANSWER: 56. (c + d)(c – d)(2c – 3d) H SOLUTION: Use the FOIL method and then combine like terms. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: ANSWER: Use the Distributive Property and then combine like 3 2 2 3 terms. 2c – 3c d – 2cd + 3d

2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 3x3 + 2x2 + x + 4 ANSWER: 3 2 5 8 3 54. (2y – 3y + 8) + (3y – 6y) 8x y z 2 –2 2 2 SOLUTION: 58. (3ab ) (2a b) SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: Use the Distributive Property and then combine like ANSWER: terms.

59. LANDSCAPING Amado wants to plant a garden ANSWER: and surround it with decorative stones. He has 2 23a – 24a enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to 56. (c + d)(c – d)(2c – 3d) cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Use the FOIL method and then combine like terms. SOLUTION: Let x be the length and y be the width of the garden.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION:

Apply the properties of exponents to simplify the expression. The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is ANSWER: 5 8 3 8x y z ANSWER: 0 to 10 ft or 24 to 34 ft 2 –2 2 2 58. (3ab ) (2a b) Solve each equation by completing the square. SOLUTION: 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

59. LANDSCAPING Amado wants to plant a garden

and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a ANSWER: perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the –3 ± width of his garden be? 2 SOLUTION: 61. x – 8x – 3 = 0 Let x be the length and y be the width of the garden. SOLUTION:

The solution of the inequality is ANSWER: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is 62. 2x2 + 6x + 5 = 0

SOLUTION: ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located. ANSWER:

–3 ± 63. SOLUTION: 2 61. x – 8x – 3 = 0 The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the SOLUTION: quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

64. ANSWER: SOLUTION: The sign of f (x) changes between the x values 2 between 0 and 1, and between 2 and 3. So the zeros 62. 2x + 6x + 5 = 0 of the quadratic function lies between 0 and 1, and SOLUTION: between 2 and 3. ANSWER: between 0 and 1; between 2 and 3

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: ANSWER: between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 State the consecutive integers between which minutes and perform a small landscape job in 90 the zeros of each quadratic function are located. minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape 63. jobs per day. Find the combination of lawns mowed SOLUTION: and completed landscape jobs per week that will The sign of f (x) changes between the x values –6 maximize income. Then find the maximum income. and –5, and between –3 and –2. So the zeros of the SOLUTION: quadratic function lies between –6 and –5, and Let M be the lawns mowed and L be the small

between –3 and –2. landscaping jobs. Since the landscaper will do at most ANSWER: 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most between –6 and –5; between –3 and –2 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he

64. can only do 11 mowing jobs per day and work a total SOLUTION: of 10 hours per day. So, for the week, if he does 15 The sign of f (x) changes between the x values landscaping jobs, he can mow 55 lawns. between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and To maximize the earnings, write a function that between 2 and 3. relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and ANSWER: $125 for each small landscaping job, F(M, L) = 35M between 0 and 1; between 2 and 3 + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

65. (M, L) F(M, L) = 35M + F(M, L) SOLUTION: 125L The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the (100, 0) F(M, L) = 35(100) + 3500 zeros of the quadratic function lies between –1 and 0, 125(0) and between 2 and 3. (55, 15) F(M, L) = 35(55) + 125 3800 ANSWER: (15) between –1 and 0; between 2 and 3 (0, 15) F(M, L) = 35(0) + 125 1875 66. BUSINESS A landscaper can mow a lawn in 30 (15) minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days 15 landscape jobs and 55 lawns; $3800 per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape ANSWER: jobs per day. Find the combination of lawns mowed 15 landscape jobs and 55 lawns; $3800 and completed landscape jobs per week that will maximize income. Then find the maximum income. 2 Find each value if f (x) = 4x + 3, g(x) = –x , and SOLUTION: h(x) = –2x2 – 2x + 4. Let M be the lawns mowed and L be the small 67. f (–6) landscaping jobs. Since the landscaper will do at most SOLUTION: 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most Substitute –6 for x in f (x). 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 ANSWER: landscaping jobs, he can mow 55 lawns. –21

To maximize the earnings, write a function that relates the number and charge for each type of job. 68. g(–8) Since he charges $35 for each lawn mowed and SOLUTION: $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and Substitute –8 for x in g(x). L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) ANSWER: 125L –64 (100, 0) F(M, L) = 35(100) + 3500 69. h(3) 125(0) SOLUTION: (55, 15) F(M, L) = 35(55) + 125 3800 Substitute 3 for x in h(x). (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800

ANSWER: ANSWER: 15 landscape jobs and 55 lawns; $3800 –20

Find each value if f (x) = 4x + 3, g(x) = –x2, and 70. f (c) h(x) = –2x2 – 2x + 4. SOLUTION: 67. f (–6) Substitute c for x in f (x).

SOLUTION:

Substitute –6 for x in f (x).

ANSWER: 4c + 3 71. g(3d) ANSWER: SOLUTION: –21 Substitute 3d for x in g(x).

68. g(–8) SOLUTION: Substitute –8 for x in g(x). ANSWER: 2 –9d

72. h(2b + 1) ANSWER: SOLUTION: –64 Substitute 2b + 1 for x in h(x). 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –8b2 – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

SOLUTION:

ANSWER: ANSWER:

4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) 5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION: SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER: ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) 2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3)

SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division. ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. ANSWER: B.

D. 2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

The correct choice is A. 10. ANSWER: A SOLUTION: Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

11.

ANSWER: SOLUTION:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: 3y + 5 Simplify

12. ANSWER: SOLUTION:

10. ANSWER: 2 2 SOLUTION: 3a b – 2ab

13.

SOLUTION:

ANSWER:

ANSWER: x + 3y –2

11. 14. SOLUTION: SOLUTION:

ANSWER:

ANSWER: 15. 3y + 5 SOLUTION: Simplify

12.

SOLUTION:

ANSWER: ANSWER: 5-2 Dividing2 Polynomials2 3a b – 2ab 2 2a + b – 3

13. 16.

SOLUTION: SOLUTION:

ANSWER: ANSWER: x + 3y –2 4c2d2 – 6

14. 17.

SOLUTION: SOLUTION:

ANSWER: ANSWER: 3np – 6 + 7p 18. ENERGY Compact fluorescent light (CFL) bulbs

15. reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 SOLUTION: be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: 2 2a + b – 3 ANSWER: –b + 8

16. 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + SOLUTION: 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. eSolutions Manual - Powered by Cognero SOLUTION: Page 4 The average number of cookies produced per worker is:

ANSWER: 4c2d2 – 6

ANSWER: 17.

SOLUTION: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER: 3np – 6 + 7p

is:

ANSWER: –b + 8 ANSWER: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by 4 3 2 –1 w to find the average number of cookies produced 22. (z – 3z + 2z – 4z + 4)(z – 1) per worker. SOLUTION: SOLUTION: The average number of cookies produced per worker is:

ANSWER: z3 – 2z2 – 4 ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER:

ANSWER: 24.

SOLUTION:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

ANSWER: 25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: z3 – 2z2 – 4 2 26. (6a – 3a + 9) ÷ (3a – 2) 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER:

27. ANSWER: SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27.

SOLUTION: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: 28.

SOLUTION: Synthetic division:

ANSWER: ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 3 2 29. (2b – 6b + 8b) ÷ (2b + 2) SOLUTION: SOLUTION:

Synthetic division:

ANSWER: ANSWER:

32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three 6 4 2 –1 dimensions always have a particular relationship 30. (6z + 3z – 9z )(3z – 6) defined by the variable x. The volume of the box can 3 2 SOLUTION: be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box?

SOLUTION: Synthetic division: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

The depressed polynomial is . ANSWER:

6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) Since the volume of the box is the product of length, SOLUTION: width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage.

SOLUTION:

ANSWER: Synthetic division:

32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? ANSWER: SOLUTION: 2 V(t) = t + 5t + 6 Divide the function by the height (x + 2) to find the length and width. Use synthetic division. 34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the The depressed polynomial is . end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number. Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 33. PHYSICS The voltage V is related to current I and 5x + 15. Divide the polynomial by x + 3. The quotient is 5. power P by the equation . The power of a ANSWER: generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If a.5 the current of the generator is I = t + 4, write an b. expression that represents the voltage. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the SOLUTION: product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

Synthetic division: sold can be estimated by ,where a is

the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? ANSWER: SOLUTION: 2 V(t) = t + 5t + 6 a. 34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the b. There are 15 hundreds in 1500. Substitute a = 15. answer. SOLUTION: a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the

x by 4 to get 4x. Then add x +15 to the product to get

5x + 15. Divide the polynomial by x + 3. The quotient Therefore, about 2423 subscriptions will be sold. is 5. ANSWER:

ANSWER: a. a.5 b. Sample answer: Let x be the number. b. about 2423 subscriptions Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + Simplify. 4 4 3. The quotient is 5. 36. (x – y ) ÷ (x – y) SOLUTION: 35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by . ANSWER: 2 2 (x + y )(x + y) b. About how many subscriptions will be sold if $1500 is spent on advertising? 3 2 2 SOLUTION: 37. (28c d – 21cd ) ÷ (14cd) SOLUTION: a.

ANSWER:

3 2 2 –1 b. There are 15 hundreds in 1500. Substitute a = 15. 38. (a b – a b + 2b)(–ab)

SOLUTION:

ANSWER:

Therefore, about 2423 subscriptions will be sold. 39. ANSWER: SOLUTION: a.

b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION: ANSWER: n2 – n – 1

40.

ANSWER: SOLUTION: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER:

41. 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION: SOLUTION:

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

39. 42. MULTIPLE REPRESENTATIONS Consider a 2 SOLUTION: rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? ANSWER: SOLUTION: 2 n – n – 1 a.

40.

SOLUTION:

ANSWER: The width is x + 3.

41. c. SOLUTION:

Yes, the concrete model checks with the algebraic model.

ANSWER: ANSWER: a.

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? The width is x + 3.

SOLUTION: b. 2 a. 2x + 7x + 3 ÷ (2x + 1)

yes

SOLUTION: b. Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3. Yes, the concrete model checks with the algebraic model. ANSWER: Sample answer: Sharon; Jamal actually divided by x ANSWER: + 3. a. 44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

The width is x + 3. 45. REASONING Review any of the division problems in this lesson. What is the relationship between the b. 2x 2 + 7x + 3 ÷ (2x + 1) degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example:

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division. In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

ANSWER: The remainder is 26. So, Sharon is correct. Jamal Sample answer: The degree of the quotient plus the actually divided by x + 3. degree of the divisor equals the degree of the dividend. ANSWER: Sample answer: Sharon; Jamal actually divided by x 46. OPEN ENDED Write a quotient of two polynomials + 3. for which the remainder is 3. SOLUTION: 44. CHALLENGE If a polynomial is divided by a Sample answer: Begin by multiplying two binomials binomial and the remainder is 0, what does this tell 2 you about the relationship between the binomial and such as (x + 2)(x + 3) which simplifies to x + 5x + the polynomial? 6. In order to get a remainder of 3 when divided, add 2 SOLUTION: 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the polynomials is . binomial and the quotient.

ANSWER: ANSWER: The binomial is a factor of the polynomial. Sample answer: 45. REASONING Review any of the division problems in this lesson. What is the relationship between the 47. CCSS ARGUMENTS Identify the expression that degrees of the dividend, the divisor, and the quotient? does not belong with the other three. Explain your SOLUTION: reasoning. Sample answer: For example:

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a polynomial.

ANSWER: In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. does not belong with the other three. The other The degree of the quotient plus the degree of the divisor equals the degree of the dividend. three expressions are polynomials. Since the denominator of contains a variable, it is not a ANSWER: Sample answer: The degree of the quotient plus the polynomial. degree of the divisor equals the degree of the dividend. 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction 46. OPEN ENDED Write a quotient of two polynomials using the division of polynomials to make a paper for which the remainder is 3. cover for your textbook. SOLUTION: SOLUTION: Sample answer: Begin by multiplying two binomials Sample answer: By dividing the area of the paper 2 such as (x + 2)(x + 3) which simplifies to x + 5x + 140x2 + 60x by the height of the book jacket 10x, the 6. In order to get a remainder of 3 when divided, add quotient of 14x + 6 provides the length of the book 2 3 to the trinomial to get x + 5x + 9. When divided, jacket. The front and back cover are each 6x units there will be a remainder of 3. The quotient of two long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is polynomials is . the width of each flap. ANSWER: ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the Sample answer: quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. 47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your 49. An office employs x women and 3 men. What is the reasoning. ratio of the total number of employees to the number of women?

SOLUTION: C does not belong with the other three. The other D three expressions are polynomials. Since the denominator of contains a variable, it is not a SOLUTION: Total number of employees: x + 3 polynomial. Number of women: x Ratio of the total number of employees to the number ANSWER: of women is: does not belong with the other three. The other

three expressions are polynomials. Since the The correct choice is A.

denominator of contains a variable, it is not a ANSWER: polynomial. A

48. WRITING IN MATH Use the information at the 50. SAT/ACT Which polynomial has degree 3? beginning of the lesson to write assembly instruction A x3 + x2 –2x4 2 using the division of polynomials to make a paper B –2x – 3x + 4 cover for your textbook. C 3x – 3 SOLUTION: D x2 + x + 123 3 Sample answer: By dividing the area of the paper E 1 + x + x 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book SOLUTION: jacket. The front and back cover are each 6x units To determine the degree of a polynomial, look at the long and the spine is 2x units long. Then, subtracting term with the greatest exponent. 14x, we are left with 6 inches. Half of this length is The correct choice is E. the width of each flap. ANSWER: ANSWER: E 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book 51. GRIDDED RESPONSE In the figure below, jacket. Then, subtracting 14x, we are left with 6 m + n + p = ? inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women?

A SOLUTION:

B Sum of the exterior angles of a triangle is . So,

C ANSWER: 360 D 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 SOLUTION: F 2x Total number of employees: x + 3 2 H –10x + 17x Number of women: x 2 Ratio of the total number of employees to the number G –10x of women is: J 2x2 + 17x

SOLUTION: The correct choice is A.

ANSWER: A The correct choice is H. 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 ANSWER: 2 H B –2x – 3x + 4 C 3x – 3 Simplify. 2 3 3 2 3 D x + x + 12 53. (5x + 2x – 3x + 4) – (2x – 4x) 3 E 1 + x + x SOLUTION: SOLUTION: Use the Distributive Property and then combine like To determine the degree of a polynomial, look at the terms. term with the greatest exponent. The correct choice is E.

ANSWER: E

51. GRIDDED RESPONSE In the figure below, ANSWER: m + n + p = ? 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

SOLUTION: Sum of the exterior angles of a triangle is . So, ANSWER: 2y 3 + 3y 2 – 9y + 8 ANSWER: 360 55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 Use the Distributive Property and then combine like F 2x terms. H –10x2 + 17x 2 G –10x 2 J 2x + 17x ANSWER: 2 SOLUTION: 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms. The correct choice is H. ANSWER: H ANSWER: 3 2 2 3 Simplify. 2c – 3c d – 2cd + 3d 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) 2 2 3 57. (xy) (2xy z) SOLUTION: Use the Distributive Property and then combine like SOLUTION: terms. Apply the properties of exponents to simplify the expression.

ANSWER: 8x5y 8z 3 ANSWER: 2 –2 2 2 3x3 + 2x2 + x + 4 58. (3ab ) (2a b) SOLUTION: 3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

ANSWER: ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: 59. LANDSCAPING Amado wants to plant a garden Use the Distributive Property and then combine like and surround it with decorative stones. He has

terms. enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? ANSWER: 2 SOLUTION: 23a – 24a Let x be the length and y be the width of the garden. 56. (c + d)(c – d)(2c – 3d)

SOLUTION: Use the FOIL method and then combine like terms.

ANSWER:

2c3 – 3c2d – 2cd 2 + 3d 3 The solution of the inequality is 2 2 3 57. (xy) (2xy z) . Since y represent the width of the garden and the sum of the length and SOLUTION: width is 34, the width of the garden is Apply the properties of exponents to simplify the expression. ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. ANSWER: 60. x2 + 6x + 2 = 0 8x5y 8z 3 SOLUTION: 2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER:

–3 ± 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has 2 enough stones to enclose a rectangular garden with a 61. x – 8x – 3 = 0 perimeter of 68 feet, but he wants the garden to SOLUTION: cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

2 62. 2x + 6x + 5 = 0 The solution of the inequality is SOLUTION: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION: ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

63.

SOLUTION:

The sign of f (x) changes between the x values –6

ANSWER: and –5, and between –3 and –2. So the zeros of the

–3 ± quadratic function lies between –6 and –5, and between –3 and –2.

2 61. x – 8x – 3 = 0 ANSWER: between –6 and –5; between –3 and –2 SOLUTION:

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and ANSWER: between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3 62. 2x2 + 6x + 5 = 0 SOLUTION: 65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 ANSWER: minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed State the consecutive integers between which and completed landscape jobs per week that will the zeros of each quadratic function are located. maximize income. Then find the maximum income. SOLUTION: 63. Let M be the lawns mowed and L be the small SOLUTION: landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L 15. Since he works – ≤ The sign of f (x) changes between the x values 6 10 hours per day, 5 days a week, he can do at most and –5, and between –3 and –2. So the zeros of the 20 mowing jobs per day or 100 per week. So, M ≤ quadratic function lies between –6 and –5, and 100.

between –3 and –2. ANSWER: If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total between –6 and –5; between –3 and –2 of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that

64. relates the number and charge for each type of job. SOLUTION: Since he charges $35 for each lawn mowed and The sign of f (x) changes between the x values $125 for each small landscaping job, F(M, L) = 35M between 0 and 1, and between 2 and 3. So the zeros + 125L. Next, substitute in the key values of M and of the quadratic function lies between 0 and 1, and L to determine the earnings for each combination of between 2 and 3. jobs.

ANSWER: (M, L) F(M, L) = 35M + F(M, L) between 0 and 1; between 2 and 3 125L

(100, 0) F(M, L) = 35(100) + 3500 125(0) 65. SOLUTION: (55, 15) F(M, L) = 35(55) + 125 3800 The sign of f (x) changes between the x values (15) between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, (0, 15) F(M, L) = 35(0) + 125 1875 and between 2 and 3. (15)

ANSWER: 15 landscape jobs and 55 lawns; $3800 between –1 and 0; between 2 and 3 ANSWER: 66. BUSINESS A landscaper can mow a lawn in 30 15 landscape jobs and 55 lawns; $3800 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days 2 Find each value if f (x) = 4x + 3, g(x) = –x , and per week. He earns $35 per lawn and $125 per 2 landscape job. He cannot do more than 3 landscape h(x) = –2x – 2x + 4. jobs per day. Find the combination of lawns mowed 67. f (–6) and completed landscape jobs per week that will SOLUTION: maximize income. Then find the maximum income. Substitute –6 for x in f (x). SOLUTION: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ ANSWER: 100. –21

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total 68. g(–8) of 10 hours per day. So, for the week, if he does 15 SOLUTION: landscaping jobs, he can mow 55 lawns. Substitute –8 for x in g(x). To maximize the earnings, write a function that relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and ANSWER: L to determine the earnings for each combination of –64 jobs. 69. h(3) (M, L) F(M, L) = 35M + F(M, L) SOLUTION: 125L Substitute 3 for x in h(x).

(100, 0) F(M, L) = 35(100) + 3500 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 ANSWER: (15) –20

15 landscape jobs and 55 lawns; $3800 70. f (c) SOLUTION: ANSWER: Substitute c for x in f (x). 15 landscape jobs and 55 lawns; $3800

2 Find each value if f (x) = 4x + 3, g(x) = –x , and h(x) = –2x2 – 2x + 4. 67. f (–6) ANSWER: SOLUTION: 4c + 3 Substitute –6 for x in f (x). 71. g(3d)

SOLUTION: Substitute 3d for x in g(x).

ANSWER: –21 ANSWER: 2 68. g(–8) –9d

SOLUTION: 72. h(2b + 1) Substitute –8 for x in g(x). SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –64 69. h(3) SOLUTION: ANSWER: Substitute 3 for x in h(x). 2 –8b – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

SOLUTION: ANSWER:

ANSWER: 4 3 2 4y + 2x – 2 5. (3z – 6z – 9z + 3z – 6) ÷ (z + 3) SOLUTION: 2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

5 2 ANSWER: 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

4 3 2 5. (3z – 6z – 9z + 3z – 6) ÷ (z + 3) ANSWER: SOLUTION: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? ANSWER:

A. B. 2 C. 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: D.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

10.

The correct choice is A. SOLUTION:

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

11.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER:

ANSWER: 3a2b – 2ab2

10.

SOLUTION: 13.

SOLUTION:

ANSWER: ANSWER: x + 3y –2

14.

11. SOLUTION:

SOLUTION:

ANSWER:

15.

SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: 2 2a + b – 3 ANSWER: 2 2 3a b – 2ab 16.

SOLUTION: 13.

SOLUTION:

ANSWER: 4c2d2 – 6

ANSWER: x + 3y –2 17.

SOLUTION: 14.

SOLUTION:

ANSWER: 3np – 6 + 7p

ANSWER: 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of 15. bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: –b + 8 ANSWER: 2 2a + b – 3 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by 16. w to find the average number of cookies produced per worker. SOLUTION: SOLUTION: The average number of cookies produced per worker is:

ANSWER: ANSWER: 4c2d2 – 6

17. Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: SOLUTION:

ANSWER: 3np – 6 + 7p ANSWER: 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of 21. (b3 – 4b2 + b – 2) ÷ (b + 1) bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION:

SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: 5-2 DividingANSWER: Polynomials –b + 8

4 3 2 –1 19. BAKING The number of cookies produced in a 22. (z – 3z + 2z – 4z + 4)(z – 1) factory each day can be estimated by –w2 + 16w + SOLUTION: 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is: ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) ANSWER: SOLUTION:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: ANSWER:

24.

ANSWER: SOLUTION:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) ANSWER: SOLUTION:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER: eSolutions Manual - Powered by Cognero Page 5

2 26. (6a – 3a + 9) ÷ (3a – 2) ANSWER: 3 2 SOLUTION: z – 2z – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

24. ANSWER: SOLUTION:

27.

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) ANSWER: SOLUTION:

28.

SOLUTION:

ANSWER: ANSWER:

3 2 27. 29. (2b – 6b + 8b) ÷ (2b + 2) SOLUTION: SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 ANSWER: 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

28. Synthetic division:

SOLUTION:

ANSWER:

ANSWER: 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1

SOLUTION:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

32. CCSS REASONING A rectangular box for a new ANSWER: product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is 6 4 2 –1 always x + 2. What are the width and length of the 30. (6z + 3z – 9z )(3z – 6) box? SOLUTION: SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

Synthetic division:

The depressed polynomial is .

ANSWER:

Since the volume of the box is the product of length, width and height, the width and length of box are (2x 6 5 3 −1 + 3)(3x + 5). 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION: ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

Synthetic division:

ANSWER:

32. CCSS REASONING A rectangular box for a new

product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 ANSWER: be written as 6x + 31x + 53x + 30, and the height is 2 always x + 2. What are the width and length of the V(t) = t + 5t + 6 box? 34. ENTERTAINMENT A magician gives these SOLUTION: instructions to a volunteer. Divide the function by the height (x + 2) to find the • Choose a number and multiply it by 4. length and width. Use synthetic division. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer.

The depressed polynomial is . SOLUTION:

a. Let x be the number.

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5). b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get ANSWER: 5x + 15. Divide the polynomial by x + 3. The quotient 2x + 3, 3x + 5 is 5.

33. PHYSICS The voltage V is related to current I and ANSWER: a.5 power P by the equation . The power of a b. Sample answer: Let x be the number. 3 2 Multiply the x by 4 to get 4x. Then add x + 15 to the generator is modeled by P(t) = t + 9t + 26t + 24. If product to get 5x + 15. Divide the polynomial by x + the current of the generator is I = t + 4, write an 3. The quotient is 5. expression that represents the voltage. SOLUTION: 35. BUSINESS The number of magazine subscriptions sold can be estimated by ,where a is

the amount of money the company spent on Synthetic division: advertising in hundreds of dollars and n is the number

of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising?

SOLUTION:

ANSWER: a. 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer.

• Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. b. There are 15 hundreds in 1500. Substitute a = 15. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

Therefore, about 2423 subscriptions will be sold.

b. Sample answer: Let x be the number. Multiply the ANSWER: x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient a. is 5. b. about 2423 subscriptions ANSWER: Simplify. a.5 4 4 b. Sample answer: Let x be the number. 36. (x – y ) ÷ (x – y) Multiply the x by 4 to get 4x. Then add x + 15 to the SOLUTION: product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on ANSWER: advertising in hundreds of dollars and n is the number 2 2 of subscriptions sold. (x + y )(x + y)

a. Perform the division indicated by . 37. (28c3d2 – 21cd2) ÷ (14cd) b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: SOLUTION:

a. ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION: b. There are 15 hundreds in 1500. Substitute a = 15.

ANSWER:

39.

Therefore, about 2423 subscriptions will be sold. SOLUTION:

ANSWER:

a. b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y) ANSWER: 2 SOLUTION: n – n – 1

40.

SOLUTION:

ANSWER: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: ANSWER:

ANSWER: 41.

SOLUTION:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER: ANSWER: 4 3 2 3z – z + 2 z – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. 39. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. SOLUTION: b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a. ANSWER: n2 – n – 1

40.

SOLUTION:

The width is x + 3.

b. ANSWER:

41.

SOLUTION: Yes, the concrete model checks with the algebraic model.

ANSWER: a.

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. The width is x + 3. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete b. 2x 2 + 7x + 3 ÷ (2x + 1) model check with your algebraic model? SOLUTION: a. c.

yes

SOLUTION: The width is x + 3. Use synthetic division.

c. The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x

Yes, the concrete model checks with the algebraic + 3. model. 44. CHALLENGE If a polynomial is divided by a ANSWER: binomial and the remainder is 0, what does this tell a. you about the relationship between the binomial and the polynomial? SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? The width is x + 3. SOLUTION: Sample answer: For example: b. 2 2x + 7x + 3 ÷ (2x + 1)

yes

reasoning. In this exercise, the degree of the dividend is 2. The SOLUTION: degree of the divisor and the quotient are each 1. Use synthetic division. The degree of the quotient plus the degree of the divisor equals the degree of the dividend. ANSWER: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend. The remainder is 26. So, Sharon is correct. Jamal 46. OPEN ENDED Write a quotient of two polynomials actually divided by x + 3. for which the remainder is 3. ANSWER: SOLUTION: Sample answer: Sharon; Jamal actually divided by x Sample answer: Begin by multiplying two binomials + 3. such as (x + 2)(x + 3) which simplifies to x2 + 5x + 44. CHALLENGE If a polynomial is divided by a 6. In order to get a remainder of 3 when divided, add 2 binomial and the remainder is 0, what does this tell 3 to the trinomial to get x + 5x + 9. When divided, you about the relationship between the binomial and there will be a remainder of 3. The quotient of two the polynomial? polynomials is . SOLUTION: If a polynomial divided by a binomial has no ANSWER: remainder, then the polynomial has two factors: the

binomial and the quotient. Sample answer: ANSWER: The binomial is a factor of the polynomial. 47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your 45. REASONING Review any of the division problems reasoning. in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient?

SOLUTION: Sample answer: For example:

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a polynomial.

ANSWER:

does not belong with the other three. The other

three expressions are polynomials. Since the In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. denominator of contains a variable, it is not a The degree of the quotient plus the degree of the divisor equals the degree of the dividend. polynomial.

ANSWER: 48. WRITING IN MATH Use the information at the Sample answer: The degree of the quotient plus the beginning of the lesson to write assembly instruction degree of the divisor equals the degree of the using the division of polynomials to make a paper dividend. cover for your textbook. SOLUTION: 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the SOLUTION: quotient of 14x + 6 provides the length of the book Sample answer: Begin by multiplying two binomials jacket. The front and back cover are each 6x units such as (x + 2)(x + 3) which simplifies to x2 + 5x + long and the spine is 2x units long. Then, subtracting 6. In order to get a remainder of 3 when divided, add 14x, we are left with 6 inches. Half of this length is 2 the width of each flap. 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two ANSWER: 2 polynomials is . Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book ANSWER: jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

Sample answer: 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number 47. CCSS ARGUMENTS Identify the expression that of women? does not belong with the other three. Explain your reasoning. A

D SOLUTION:

does not belong with the other three. The other SOLUTION: Total number of employees: x + 3 three expressions are polynomials. Since the Number of women: x denominator of contains a variable, it is not a Ratio of the total number of employees to the number of women is: polynomial. ANSWER: The correct choice is A. does not belong with the other three. The other ANSWER: three expressions are polynomials. Since the A

denominator of contains a variable, it is not a 50. SAT/ACT Which polynomial has degree 3? 3 2 4 polynomial. A x + x –2x 2 B –2x – 3x + 4 48. WRITING IN MATH Use the information at the C 3x – 3 beginning of the lesson to write assembly instruction D 2 3 using the division of polynomials to make a paper x + x + 12 3 cover for your textbook. E 1 + x + x SOLUTION: SOLUTION: Sample answer: By dividing the area of the paper To determine the degree of a polynomial, look at the 140x2 + 60x by the height of the book jacket 10x, the term with the greatest exponent. quotient of 14x + 6 provides the length of the book The correct choice is E. jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting ANSWER: 14x, we are left with 6 inches. Half of this length is E the width of each flap. 51. GRIDDED RESPONSE In the figure below, ANSWER: m + n + p = ? 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number SOLUTION: of women? Sum of the exterior angles of a triangle is .

A So, ANSWER: B 360

C 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x D H –10x2 + 17x 2 SOLUTION: G –10x Total number of employees: x + 3 J 2x2 + 17x Number of women: x Ratio of the total number of employees to the number SOLUTION: of women is:

The correct choice is A. The correct choice is H. ANSWER: A ANSWER: H 50. SAT/ACT Which polynomial has degree 3? 3 2 4 Simplify. A x + x –2x 3 2 3 2 53. (5x + 2x – 3x + 4) – (2x – 4x) B –2x – 3x + 4 C 3x – 3 SOLUTION: D x2 + x + 123 Use the Distributive Property and then combine like 3 terms. E 1 + x + x

SOLUTION: To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is E.

ANSWER: ANSWER: E 3x3 + 2x2 + x + 4 51. GRIDDED RESPONSE In the figure below, m + n + p = ? 3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

ANSWER: SOLUTION: 3 2 2y + 3y – 9y + 8 Sum of the exterior angles of a triangle is . So, 55. 4a(2a – 3) + 3a(5a – 4)

ANSWER: SOLUTION: 360 Use the Distributive Property and then combine like terms. 2 2 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2 F 2x H –10x2 + 17x ANSWER: 2 2 G –10x 23a – 24a J 2x2 + 17x 56. (c + d)(c – d)(2c – 3d) SOLUTION: SOLUTION: Use the FOIL method and then combine like terms.

The correct choice is H. ANSWER: 3 2 2 3 ANSWER: 2c – 3c d – 2cd + 3d H 2 2 3 57. (xy) (2xy z) Simplify. SOLUTION: 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) Apply the properties of exponents to simplify the SOLUTION: expression. Use the Distributive Property and then combine like terms.

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b)

ANSWER: SOLUTION: 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION: ANSWER:

ANSWER: 2y 3 + 3y 2 – 9y + 8 59. LANDSCAPING Amado wants to plant a garden 55. 4a(2a – 3) + 3a(5a – 4) and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a SOLUTION: perimeter of 68 feet, but he wants the garden to Use the Distributive Property and then combine like cover no more than 240 square feet. What could the terms. width of his garden be?

SOLUTION: Let x be the length and y be the width of the garden. ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms.

The solution of the inequality is ANSWER: . Since y represent the 2c3 – 3c2d – 2cd 2 + 3d 3 width of the garden and the sum of the length and width is 34, the width of the garden is 2 2 3 57. (xy) (2xy z) SOLUTION: ANSWER: Apply the properties of exponents to simplify the 0 to 10 ft or 24 to 34 ft expression. Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION: ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER: ANSWER: –3 ±

2 61. x – 8x – 3 = 0 59. LANDSCAPING Amado wants to plant a garden SOLUTION: and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

62. 2x2 + 6x + 5 = 0 SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 ANSWER: SOLUTION:

State the consecutive integers between which the zeros of each quadratic function are located.

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2. ANSWER: ANSWER: –3 ± between –6 and –5; between –3 and –2

2 61. x – 8x – 3 = 0 SOLUTION: 64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: ANSWER: between 0 and 1; between 2 and 3

62. 2x2 + 6x + 5 = 0 65. SOLUTION: SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape ANSWER: jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. SOLUTION: State the consecutive integers between which the zeros of each quadratic function are located. Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 63. 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M SOLUTION: ≤ 100. The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the If the landscaper does 3 landscaping jobs per day, he quadratic function lies between –6 and –5, and can only do 11 mowing jobs per day and work a total between –3 and –2. of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. ANSWER:

between –6 and –5; between –3 and –2 To maximize the earnings, write a function that relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M 64. + 125L. Next, substitute in the key values of M and SOLUTION: L to determine the earnings for each combination of jobs. The sign of f (x) changes between the x values

between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and (M, L) F(M, L) = 35M + F(M, L) between 2 and 3. 125L

ANSWER: (100, 0) F(M, L) = 35(100) + 3500 between 0 and 1; between 2 and 3 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 (15) 65. SOLUTION: (0, 15) F(M, L) = 35(0) + 125 1875 The sign of f (x) changes between the x values (15) between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, 15 landscape jobs and 55 lawns; $3800 and between 2 and 3. ANSWER: ANSWER: 15 landscape jobs and 55 lawns; $3800 between –1 and 0; between 2 and 3 Find each value if f (x) = 4x + 3, g(x) = –x2, and 66. BUSINESS A landscaper can mow a lawn in 30 2 minutes and perform a small landscape job in 90 h(x) = –2x – 2x + 4. minutes. He works at most 10 hours per day, 5 days 67. f (–6) per week. He earns $35 per lawn and $125 per SOLUTION: landscape job. He cannot do more than 3 landscape Substitute –6 for x in f (x). jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. SOLUTION: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most ANSWER: 3 landscaping jobs per day, L ≤ 15. Since he works –21 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100. 68. g(–8) SOLUTION: If the landscaper does 3 landscaping jobs per day, he Substitute –8 for x in g(x). can only do 11 mowing jobs per day and work a total

of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that relates the number and charge for each type of job. ANSWER: Since he charges $35 for each lawn mowed and –64 $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and 69. h(3) L to determine the earnings for each combination of SOLUTION: jobs. Substitute 3 for x in h(x).

(M, L) F(M, L) = 35M + F(M, L) 125L

(100, 0) F(M, L) = 35(100) + 3500 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 ANSWER: (15) –20 (0, 15) F(M, L) = 35(0) + 125 1875 70. f (c) (15) SOLUTION: Substitute c for x in f (x). 15 landscape jobs and 55 lawns; $3800 ANSWER: 15 landscape jobs and 55 lawns; $3800 ANSWER: Find each value if f (x) = 4x + 3, g(x) = –x2, and 4c + 3 h(x) = –2x2 – 2x + 4. 67. f (–6) 71. g(3d) SOLUTION: SOLUTION: Substitute –6 for x in f (x). Substitute 3d for x in g(x).

ANSWER: ANSWER: 2 –9d –21 72. h(2b + 1) 68. g(–8) SOLUTION: SOLUTION: Substitute 2b + 1 for x in h(x). Substitute –8 for x in g(x).

ANSWER: –64 ANSWER: 69. h(3) –8b2 – 12b SOLUTION: Substitute 3 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

Simplify. 2 1. 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION: ANSWER:

ANSWER: 5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) 3a + 5b – 6 SOLUTION: 3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) ANSWER: SOLUTION:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A . B.

C. ANSWER:

5 2 SOLUTION: 6. (y – 3y – 20) ÷ (y – 2) First rewrite the divisor so the x-term is first. Then SOLUTION: use long division.

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) ANSWER: SOLUTION: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: ANSWER: First rewrite the divisor so the x-term is first. Then use long division.

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

The correct choice is A.

ANSWER: ANSWER: A

Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5)

SOLUTION: 10.

SOLUTION:

ANSWER: ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) 11. SOLUTION: SOLUTION:

ANSWER: ANSWER: 3y + 5

Simplify 10. 12. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13.

ANSWER: SOLUTION:

11.

SOLUTION: ANSWER: x + 3y –2

14.

SOLUTION:

ANSWER: 3y + 5 Simplify ANSWER:

12.

SOLUTION: 15.

SOLUTION:

ANSWER: 3a2b – 2ab2

13.

SOLUTION: ANSWER: 2 2a + b – 3

16.

SOLUTION:

ANSWER: x + 3y –2

14.

SOLUTION: ANSWER: 4c2d2 – 6

17.

SOLUTION: ANSWER:

15.

SOLUTION: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. ANSWER: 2 SOLUTION: 2a + b – 3 The average amount of energy saved per CFL bulb is: 16.

SOLUTION:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced ANSWER: per worker. 4c2d2 – 6 SOLUTION: The average number of cookies produced per worker 17. is:

SOLUTION:

ANSWER:

ANSWER: Simplify. 3np – 6 + 7p 2 20. (a – 8a – 26) ÷ (a + 2) 18. ENERGY Compact fluorescent light (CFL) bulbs SOLUTION: reduce energy waste. The amount of energy waste

that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb ANSWER: is:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) ANSWER: –b + 8 SOLUTION:

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: ANSWER: The average number of cookies produced per worker is:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

Simplify. 2 ANSWER: 20. (a – 8a – 26) ÷ (a + 2) z3 – 2z2 – 4 SOLUTION: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: 24.

SOLUTION:

ANSWER:

ANSWER: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) ANSWER: SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER: ANSWER:

5-2 Dividing Polynomials

25. (g4 – 3g2 – 18) ÷ (g – 2) 27. SOLUTION: SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27.

SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: eSolutions Manual - Powered by Cognero Page 6

Synthetic division:

ANSWER:

28. ANSWER:

SOLUTION:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) ANSWER: SOLUTION:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) ANSWER: SOLUTION:

32. CCSS REASONING A rectangular box for a new

product is designed in such a way that the three

Synthetic division: dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division. ANSWER:

6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION: The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a 3 2 ANSWER: generator is modeled by P(t) = t + 9t + 26t + 24. If the current of the generator is I = t + 4, write an

expression that represents the voltage.

32. CCSS REASONING A rectangular box for a new SOLUTION: product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is Synthetic division: always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER: 2 V(t) = t + 5t + 6

The depressed polynomial is . 34. ENTERTAINMENT A magician gives these instructions to a volunteer.

• Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. Since the volume of the box is the product of length, a. What number will the volunteer always have at the width and height, the width and length of box are (2x end? + 3)(3x + 5). b. Explain the process you used to discover the answer. ANSWER: SOLUTION: 2x + 3, 3x + 5 a. Let x be the number.

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an b. Sample answer: Let x be the number. Multiply the

expression that represents the voltage. x by 4 to get 4x. Then add x +15 to the product to get SOLUTION: 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 Synthetic division: b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on ANSWER: advertising in hundreds of dollars and n is the number 2 V(t) = t + 5t + 6 of subscriptions sold.

34. ENTERTAINMENT A magician gives these a. Perform the division indicated by . instructions to a volunteer. b. About how many subscriptions will be sold if • Choose a number and multiply it by 4. $1500 is spent on advertising? • Then add the sum of your number and 15 to the product you found. SOLUTION: • Now divide by the sum of your number and 3. a . What number will the volunteer always have at the a. end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

b. There are 15 hundreds in 1500. Substitute a = 15.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the Therefore, about 2423 subscriptions will be sold. product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. ANSWER:

35. BUSINESS The number of magazine subscriptions a. sold can be estimated by ,where a is b. about 2423 subscriptions the amount of money the company spent on Simplify. advertising in hundreds of dollars and n is the number 36. (x4 – y4) ÷ (x – y) of subscriptions sold. SOLUTION: a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION:

a. ANSWER: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd)

SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15.

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER: ANSWER: a. b. about 2423 subscriptions

Simplify. 39. 36. (x4 – y4) ÷ (x – y) SOLUTION: SOLUTION:

ANSWER: 2 2 ANSWER: (x + y )(x + y) n2 – n – 1

37. (28c3d2 – 21cd2) ÷ (14cd) 40. SOLUTION: SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) ANSWER:

SOLUTION:

41.

ANSWER: SOLUTION:

39.

SOLUTION: ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. ANSWER: a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. n2 – n – 1 b. SYMBOLIC Write an expression to represent

the model. 40. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: SOLUTION: a.

ANSWER:

41. The width is x + 3. SOLUTION: b.

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9– Yes, the concrete model checks with the algebraic model. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. ANSWER: a. CONCRETE Use algebra tiles to represent this a. situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

The width is x + 3. yes b. 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the

remainder is 26. Jamal argues that the remainder is – c. 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

Yes, the concrete model checks with the algebraic model.

ANSWER: a. The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial?

SOLUTION: The width is x + 3. If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the b. 2x 2 + 7x + 3 ÷ (2x + 1) binomial and the quotient.

ANSWER:

The binomial is a factor of the polynomial. c. 45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? yes SOLUTION: 43. ERROR ANALYSIS Sharon and Jamal are dividing Sample answer: For example: 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3. In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. ANSWER: The degree of the quotient plus the degree of the Sample answer: Sharon; Jamal actually divided by x divisor equals the degree of the dividend. + 3. ANSWER: 44. CHALLENGE If a polynomial is divided by a Sample answer: The degree of the quotient plus the binomial and the remainder is 0, what does this tell degree of the divisor equals the degree of the you about the relationship between the binomial and dividend. the polynomial? 46. OPEN ENDED Write a quotient of two polynomials SOLUTION: for which the remainder is 3. If a polynomial divided by a binomial has no SOLUTION: remainder, then the polynomial has two factors: the binomial and the quotient. Sample answer: Begin by multiplying two binomials such as (x + 2)(x + 3) which simplifies to x2 + 5x + ANSWER: 6. In order to get a remainder of 3 when divided, add 2 The binomial is a factor of the polynomial. 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two 45. REASONING Review any of the division problems

in this lesson. What is the relationship between the polynomials is . degrees of the dividend, the divisor, and the quotient? SOLUTION: ANSWER: Sample answer: For example: Sample answer:

47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

SOLUTION:

does not belong with the other three. The other In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. three expressions are polynomials. Since the The degree of the quotient plus the degree of the divisor equals the degree of the dividend. denominator of contains a variable, it is not a polynomial. ANSWER: Sample answer: The degree of the quotient plus the ANSWER: degree of the divisor equals the degree of the dividend. does not belong with the other three. The other

46. OPEN ENDED Write a quotient of two polynomials three expressions are polynomials. Since the for which the remainder is 3. denominator of contains a variable, it is not a SOLUTION: polynomial. Sample answer: Begin by multiplying two binomials such as (x + 2)(x + 3) which simplifies to x2 + 5x + 48. WRITING IN MATH Use the information at the 6. In order to get a remainder of 3 when divided, add beginning of the lesson to write assembly instruction 2 3 to the trinomial to get x + 5x + 9. When divided, using the division of polynomials to make a paper

there will be a remainder of 3. The quotient of two cover for your textbook. SOLUTION: polynomials is . Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the ANSWER: quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units Sample answer: long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is 47. CCSS ARGUMENTS Identify the expression that the width of each flap. does not belong with the other three. Explain your reasoning. ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book

jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the SOLUTION: ratio of the total number of employees to the number of women? does not belong with the other three. The other A three expressions are polynomials. Since the

denominator of contains a variable, it is not a B

polynomial. C ANSWER: D does not belong with the other three. The other three expressions are polynomials. Since the SOLUTION: Total number of employees: x + 3 denominator of contains a variable, it is not a Number of women: x Ratio of the total number of employees to the number polynomial. of women is: 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction using the division of polynomials to make a paper The correct choice is A. cover for your textbook. ANSWER: SOLUTION: A Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the 50. SAT/ACT Which polynomial has degree 3? 3 2 4 quotient of 14x + 6 provides the length of the book A x + x –2x 2 jacket. The front and back cover are each 6x units B –2x – 3x + 4 long and the spine is 2x units long. Then, subtracting C 3x – 3 14x, we are left with 6 inches. Half of this length is 2 3 the width of each flap. D x + x + 12 3 E 1 + x + x ANSWER: 2 SOLUTION: Sample answer: By dividing 140x + 60x by 10x, the To determine the degree of a polynomial, look at the quotient of 14x + 6 provides the length of the book

jacket. Then, subtracting 14x, we are left with 6 term with the greatest exponent. inches. Half of this length is the width of each flap. The correct choice is E.

49. An office employs x women and 3 men. What is the ANSWER: ratio of the total number of employees to the number E of women? 51. GRIDDED RESPONSE In the figure below, A m + n + p = ?

D SOLUTION: SOLUTION: Sum of the exterior angles of a triangle is . Total number of employees: x + 3 So, Number of women: x ANSWER: Ratio of the total number of employees to the number of women is: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 The correct choice is A. F 2x H –10x2 + 17x ANSWER: G 2 A –10x J 2x2 + 17x 50. SAT/ACT Which polynomial has degree 3? SOLUTION: A x3 + x2 –2x4 2 B –2x – 3x + 4 C 3x – 3 D x2 + x + 123 3 E 1 + x + x The correct choice is H. SOLUTION: ANSWER: To determine the degree of a polynomial, look at the H term with the greatest exponent. Simplify. The correct choice is E. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) ANSWER: SOLUTION: E Use the Distributive Property and then combine like 51. GRIDDED RESPONSE In the figure below, terms. m + n + p = ?

ANSWER: SOLUTION: 3x3 + 2x2 + x + 4 Sum of the exterior angles of a triangle is . So, 3 2 54. (2y – 3y + 8) + (3y – 6y) ANSWER: SOLUTION: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x ANSWER: 2 H –10x + 17x 2y 3 + 3y 2 – 9y + 8 2 G –10x 2 55. 4a(2a – 3) + 3a(5a – 4) J 2x + 17x SOLUTION: SOLUTION: Use the Distributive Property and then combine like terms.

The correct choice is H. ANSWER: 2 ANSWER: 23a – 24a H 56. (c + d)(c – d)(2c – 3d) Simplify. SOLUTION: 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) Use the FOIL method and then combine like terms. SOLUTION: Use the Distributive Property and then combine like terms. ANSWER:

3 2 2 3 2c – 3c d – 2cd + 3d 2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the ANSWER: expression.

3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) ANSWER: SOLUTION: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b)

ANSWER: SOLUTION: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: Use the Distributive Property and then combine like terms. ANSWER:

ANSWER: 2 23a – 24a 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has 56. (c + d)(c – d)(2c – 3d) enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to SOLUTION: cover no more than 240 square feet. What could the Use the FOIL method and then combine like terms. width of his garden be?

SOLUTION: Let x be the length and y be the width of the garden. ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the expression.

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and ANSWER: width is 34, the width of the garden is 5 8 3 8x y z

2 –2 2 2 58. (3ab ) (2a b) ANSWER: 0 to 10 ft or 24 to 34 ft SOLUTION: Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a

perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the ANSWER: width of his garden be? –3 ± SOLUTION: Let x be the length and y be the width of the garden. 2 61. x – 8x – 3 = 0

SOLUTION:

The solution of the inequality is . Since y represent the ANSWER: width of the garden and the sum of the length and

width is 34, the width of the garden is

62. 2x2 + 6x + 5 = 0 ANSWER: SOLUTION: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

ANSWER: State the consecutive integers between which the zeros of each quadratic function are located. –3 ±

63. 2 61. x – 8x – 3 = 0 SOLUTION: SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

ANSWER: 64. SOLUTION: 2 62. 2x + 6x + 5 = 0 The sign of f (x) changes between the x values SOLUTION: between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, ANSWER: and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3 State the consecutive integers between which the zeros of each quadratic function are located. 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days

63. per week. He earns $35 per lawn and $125 per SOLUTION: landscape job. He cannot do more than 3 landscape The sign of f (x) changes between the x values –6 jobs per day. Find the combination of lawns mowed and –5, and between –3 and –2. So the zeros of the and completed landscape jobs per week that will

quadratic function lies between –6 and –5, and maximize income. Then find the maximum income. between –3 and –2. SOLUTION: ANSWER: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most between –6 and –5; between –3 and –2 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100. 64. SOLUTION: If the landscaper does 3 landscaping jobs per day, he The sign of f (x) changes between the x values can only do 11 mowing jobs per day and work a total between 0 and 1, and between 2 and 3. So the zeros of 10 hours per day. So, for the week, if he does 15 of the quadratic function lies between 0 and 1, and landscaping jobs, he can mow 55 lawns. between 2 and 3. To maximize the earnings, write a function that ANSWER: relates the number and charge for each type of job. between 0 and 1; between 2 and 3 Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of 65. jobs.

SOLUTION: The sign of f (x) changes between the x values (M, L) F(M, L) = 35M + F(M, L) between –1 and 0, and between 2 and 3. So the 125L zeros of the quadratic function lies between –1 and 0, and between 2 and 3. (100, 0) F(M, L) = 35(100) + 3500 125(0) ANSWER: between –1 and 0; between 2 and 3 (55, 15) F(M, L) = 35(55) + 125 3800 (15) 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 (0, 15) F(M, L) = 35(0) + 125 1875 minutes. He works at most 10 hours per day, 5 days (15) per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape 15 landscape jobs and 55 lawns; $3800 jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will ANSWER: maximize income. Then find the maximum income. 15 landscape jobs and 55 lawns; $3800

SOLUTION: 2 Let M be the lawns mowed and L be the small Find each value if f (x) = 4x + 3, g(x) = –x , and landscaping jobs. Since the landscaper will do at most h(x) = –2x2 – 2x + 4. 3 landscaping jobs per day, L ≤ 15. Since he works 67. f (–6) 10 hours per day, 5 days a week, he can do at most SOLUTION: 20 mowing jobs per day or 100 per week. So, M ≤ 100. Substitute –6 for x in f (x).

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. ANSWER: To maximize the earnings, write a function that –21 relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M 68. g(–8) + 125L. Next, substitute in the key values of M and SOLUTION: L to determine the earnings for each combination of jobs. Substitute –8 for x in g(x).

(M, L) F(M, L) = 35M + F(M, L) 125L ANSWER: (100, 0) F(M, L) = 35(100) + 3500 –64 125(0) 69. h(3) (55, 15) F(M, L) = 35(55) + 125 3800 (15) SOLUTION: Substitute 3 for x in h(x). (0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800

ANSWER: 15 landscape jobs and 55 lawns; $3800 ANSWER: Find each value if f (x) = 4x + 3, g(x) = –x2, and –20 2 h(x) = –2x – 2x + 4. 70. f (c) 67. f (–6) SOLUTION: SOLUTION: Substitute c for x in f (x). Substitute –6 for x in f (x).

ANSWER: 4c + 3 ANSWER: –21 71. g(3d) SOLUTION: 68. g(–8) Substitute 3d for x in g(x).

SOLUTION:

Substitute –8 for x in g(x).

ANSWER: 2 –9d ANSWER: –64 72. h(2b + 1) SOLUTION: 69. h(3) Substitute 2b + 1 for x in h(x). SOLUTION: Substitute 3 for x in h(x).

ANSWER: 2 ANSWER: –8b – 12b –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

SOLUTION:

ANSWER: ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) 4 3 2 SOLUTION: 5. (3z – 6z – 9z + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER: ANSWER:

2 5 2 4. (2a – 4a – 8) ÷ (a + 1) 6. (y – 3y – 20) ÷ (y – 2) SOLUTION: SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3)

SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then ANSWER: use long division.

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B. ANSWER: C.

D. 2 SOLUTION: 9. (18a + 6a + 9) ÷ (3a – 2) First rewrite the divisor so the x-term is first. Then SOLUTION: use long division.

ANSWER:

The correct choice is A.

ANSWER: 10. A SOLUTION: Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

11. ANSWER:

SOLUTION:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: 3y + 5 Simplify

ANSWER: 12.

SOLUTION:

10. ANSWER: SOLUTION: 3a2b – 2ab2

13.

SOLUTION:

ANSWER:

ANSWER: x + 3y –2 11.

SOLUTION: 14.

SOLUTION:

ANSWER:

3y + 5 15. Simplify SOLUTION: 12.

SOLUTION:

ANSWER: 3a2b – 2ab2 ANSWER: 2 2a + b – 3 13.

16. SOLUTION: SOLUTION:

ANSWER: x + 3y –2 ANSWER: 2 2 4c d – 6 14.

17. SOLUTION: SOLUTION:

ANSWER: ANSWER: 3np – 6 + 7p

15. 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste SOLUTION: that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: 2 2a + b – 3 ANSWER:

16. –b + 8 19. BAKING The number of cookies produced in a SOLUTION: factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is: ANSWER: 4c2d2 – 6

17. ANSWER:

SOLUTION:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of ANSWER: bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb 21. (b3 – 4b2 + b – 2) ÷ (b + 1) is: SOLUTION:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a ANSWER: factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION: SOLUTION: The average number of cookies produced per worker

is:

ANSWER: ANSWER: 3 2 z – 2z – 4 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) Simplify. SOLUTION: 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER:

24.

SOLUTION: 3 2 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

SOLUTION:

ANSWER: ANSWER: z3 – 2z2 – 4

5 3 2 2 23. (x – 4x + 4x ) ÷ (x – 4) 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION: SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER:

ANSWER: 27.

SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2)

SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

ANSWER: 27.

SOLUTION:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER: ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) 28. SOLUTION: SOLUTION:

Synthetic division:

ANSWER: ANSWER:

5-2 Dividing Polynomials

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION: SOLUTION:

Synthetic division:

ANSWER: ANSWER:

32. CCSS REASONING A rectangular box for a new 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) product is designed in such a way that the three dimensions always have a particular relationship SOLUTION: defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? Synthetic division: SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER: The depressed polynomial is .

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1

SOLUTION: Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 eSolutions Manual - Powered by Cognero Page 7 33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

ANSWER:

Synthetic division: 32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box?

SOLUTION: ANSWER: Divide the function by the height (x + 2) to find the 2 length and width. Use synthetic division. V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these

instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found.

• Now divide by the sum of your number and 3. The depressed polynomial is . a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION: Since the volume of the box is the product of length, a. width and height, the width and length of box are (2x Let x be the number. + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 b. Sample answer: Let x be the number. Multiply the PHYSICS 33. The voltage V is related to current I and x by 4 to get 4x. Then add x +15 to the product to get power P by the equation . The power of a 5x + 15. Divide the polynomial by x + 3. The quotient is 5. generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an ANSWER: expression that represents the voltage. a.5 b. Sample answer: Let x be the number. SOLUTION: Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

BUSINESS Synthetic division: 35. The number of magazine subscriptions sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by . b. About how many subscriptions will be sold if ANSWER: $1500 is spent on advertising? 2 V(t) = t + 5t + 6 SOLUTION: 34. ENTERTAINMENT A magician gives these instructions to a volunteer. a. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b . Explain the process you used to discover the answer. b. There are 15 hundreds in 1500. Substitute a = 15. SOLUTION: a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. Therefore, about 2423 subscriptions will be sold.

ANSWER: ANSWER: a.5 a. b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the b. about 2423 subscriptions product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. Simplify. 36. (x4 – y4) ÷ (x – y) 35. BUSINESS The number of magazine subscriptions SOLUTION: sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by . ANSWER: b. About how many subscriptions will be sold if 2 2 $1500 is spent on advertising? (x + y )(x + y) SOLUTION: 3 2 2 37. (28c d – 21cd ) ÷ (14cd) a. SOLUTION:

ANSWER:

b. There are 15 hundreds in 1500. Substitute a = 15. 3 2 2 –1 38. (a b – a b + 2b)(–ab)

SOLUTION:

ANSWER:

Therefore, about 2423 subscriptions will be sold.

ANSWER: 39.

a. SOLUTION: b. about 2423 subscriptions Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION:

ANSWER: n2 – n – 1

40. ANSWER: 2 2 (x + y )(x + y) SOLUTION:

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER: ANSWER:

3 2 2 –1 41. 38. (a b – a b + 2b)(–ab) SOLUTION: SOLUTION:

ANSWER:

ANSWER: 4 3 2 39. 3z – z + 2 z – 4z + 9–

SOLUTION: 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete ANSWER: model check with your algebraic model? n2 – n – 1 SOLUTION: a. 40.

SOLUTION:

ANSWER: The width is x + 3.

41.

SOLUTION: c.

Yes, the concrete model checks with the algebraic model.

ANSWER: ANSWER: a. 3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: The width is x + 3.

a. b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the The width is x + 3. remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. b. SOLUTION: Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal Yes, the concrete model checks with the algebraic actually divided by x + 3. model. ANSWER: ANSWER: Sample answer: Sharon; Jamal actually divided by x a. + 3.

ANSWER: The width is x + 3. The binomial is a factor of the polynomial.

REASONING 2 45. Review any of the division problems b. 2x + 7x + 3 ÷ (2x + 1) in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient?

SOLUTION: c. Sample answer: For example:

yes

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

The remainder is 26. So, Sharon is correct. Jamal ANSWER: actually divided by x + 3. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the ANSWER: dividend. Sample answer: Sharon; Jamal actually divided by x + 3. 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. 44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell SOLUTION: you about the relationship between the binomial and Sample answer: Begin by multiplying two binomials the polynomial? such as (x + 2)(x + 3) which simplifies to x2 + 5x + SOLUTION: 6. In order to get a remainder of 3 when divided, add 2 If a polynomial divided by a binomial has no 3 to the trinomial to get x + 5x + 9. When divided, remainder, then the polynomial has two factors: the there will be a remainder of 3. The quotient of two binomial and the quotient. polynomials is . ANSWER: The binomial is a factor of the polynomial. ANSWER:

45. REASONING Review any of the division problems Sample answer: in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? 47. CCSS ARGUMENTS Identify the expression that SOLUTION: does not belong with the other three. Explain your Sample answer: For example: reasoning.

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a

polynomial. In this exercise, the degree of the dividend is 2. The ANSWER: degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the does not belong with the other three. The other divisor equals the degree of the dividend. three expressions are polynomials. Since the ANSWER: Sample answer: The degree of the quotient plus the denominator of contains a variable, it is not a degree of the divisor equals the degree of the polynomial. dividend. 48. WRITING IN MATH Use the information at the OPEN ENDED 46. Write a quotient of two polynomials beginning of the lesson to write assembly instruction

for which the remainder is 3. using the division of polynomials to make a paper SOLUTION: cover for your textbook. Sample answer: Begin by multiplying two binomials SOLUTION: 2 such as (x + 2)(x + 3) which simplifies to x + 5x + Sample answer: By dividing the area of the paper 6. In order to get a remainder of 3 when divided, add 140x2 + 60x by the height of the book jacket 10x, the 2 3 to the trinomial to get x + 5x + 9. When divided, quotient of 14x + 6 provides the length of the book there will be a remainder of 3. The quotient of two jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting polynomials is . 14x, we are left with 6 inches. Half of this length is the width of each flap. ANSWER: ANSWER:

2 Sample answer: Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book 47. CCSS ARGUMENTS Identify the expression that jacket. Then, subtracting 14x, we are left with 6 does not belong with the other three. Explain your inches. Half of this length is the width of each flap. reasoning. 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women?

B SOLUTION:

does not belong with the other three. The other C three expressions are polynomials. Since the D denominator of contains a variable, it is not a SOLUTION: polynomial. Total number of employees: x + 3 ANSWER: Number of women: x Ratio of the total number of employees to the number does not belong with the other three. The other of women is:

three expressions are polynomials. Since the denominator of contains a variable, it is not a The correct choice is A. polynomial. ANSWER: A 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction 50. SAT/ACT Which polynomial has degree 3? using the division of polynomials to make a paper A x3 + x2 –2x4 cover for your textbook. 2 B –2x – 3x + 4 SOLUTION: C 3x – 3 Sample answer: By dividing the area of the paper D x2 + x + 123 2 3 140x + 60x by the height of the book jacket 10x, the E 1 + x + x quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units SOLUTION: long and the spine is 2x units long. Then, subtracting To determine the degree of a polynomial, look at the 14x, we are left with 6 inches. Half of this length is term with the greatest exponent. the width of each flap. The correct choice is E. ANSWER: ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the E quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 51. GRIDDED RESPONSE In the figure below, inches. Half of this length is the width of each flap. m + n + p = ? 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women?

B SOLUTION: Sum of the exterior angles of a triangle is .

C So, ANSWER: D 360

SOLUTION: 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = Total number of employees: x + 3 2 F Number of women: x 2x 2 Ratio of the total number of employees to the number H –10x + 17x 2 of women is: G –10x 2 J 2x + 17x The correct choice is A. SOLUTION:

ANSWER: A

50. SAT/ACT Which polynomial has degree 3? The correct choice is H. A x3 + x2 –2x4 2 B –2x – 3x + 4 ANSWER: C 3x – 3 H 2 3 D x + x + 12 Simplify. 3 3 2 3 E 1 + x + x 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: SOLUTION: To determine the degree of a polynomial, look at the Use the Distributive Property and then combine like term with the greatest exponent. terms. The correct choice is E.

ANSWER: E

51. GRIDDED RESPONSE In the figure below, m + n + p = ? ANSWER: 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y)

SOLUTION: SOLUTION: Sum of the exterior angles of a triangle is . So, ANSWER: ANSWER: 2y 3 + 3y 2 – 9y + 8 360 55. 4a(2a – 3) + 3a(5a – 4) 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 SOLUTION: F 2x 2 Use the Distributive Property and then combine like H –10x + 17x terms. 2 G –10x J 2x2 + 17x SOLUTION: ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d)

SOLUTION: The correct choice is H. Use the FOIL method and then combine like terms. ANSWER: H Simplify. ANSWER: 3 2 3 3 2 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) 2c – 3c d – 2cd + 3d

SOLUTION: 2 2 3 57. (xy) (2xy z) Use the Distributive Property and then combine like terms. SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: ANSWER: 8x5y 8z 3 3x3 + 2x2 + x + 4 2 –2 2 2 58. (3ab ) (2a b) 3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION: SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8 ANSWER: 55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: Use the Distributive Property and then combine like terms. 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to ANSWER: cover no more than 240 square feet. What could the 2 width of his garden be? 23a – 24a SOLUTION: 56. (c + d)(c – d)(2c – 3d) Let x be the length and y be the width of the garden. SOLUTION: Use the FOIL method and then combine like terms.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) The solution of the inequality is . Since y represent the SOLUTION: width of the garden and the sum of the length and Apply the properties of exponents to simplify the width is 34, the width of the garden is expression.

ANSWER: 0 to 10 ft or 24 to 34 ft ANSWER: Solve each equation by completing the square. 8x5y 8z 3 60. x2 + 6x + 2 = 0 2 –2 2 2 58. (3ab ) (2a b) SOLUTION: SOLUTION:

ANSWER:

ANSWER: LANDSCAPING 59. Amado wants to plant a garden –3 ± and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a 2 perimeter of 68 feet, but he wants the garden to 61. x – 8x – 3 = 0 cover no more than 240 square feet. What could the width of his garden be? SOLUTION: SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

2 The solution of the inequality is 62. 2x + 6x + 5 = 0 . Since y represent the SOLUTION: width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

63.

SOLUTION: ANSWER: The sign of f (x) changes between the x values –6 –3 ± and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and 2 between –3 and –2. 61. x – 8x – 3 = 0 ANSWER: SOLUTION: between –6 and –5; between –3 and –2

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros ANSWER: of the quadratic function lies between 0 and 1, and between 2 and 3.

2 ANSWER: 62. 2x + 6x + 5 = 0 between 0 and 1; between 2 and 3 SOLUTION:

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

ANSWER: 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape State the consecutive integers between which jobs per day. Find the combination of lawns mowed the zeros of each quadratic function are located. and completed landscape jobs per week that will maximize income. Then find the maximum income.

63. SOLUTION: SOLUTION: Let M be the lawns mowed and L be the small The sign of f (x) changes between the x values –6 landscaping jobs. Since the landscaper will do at most and –5, and between –3 and –2. So the zeros of the 3 landscaping jobs per day, L ≤ 15. Since he works quadratic function lies between –6 and –5, and 10 hours per day, 5 days a week, he can do at most between –3 and –2. 20 mowing jobs per day or 100 per week. So, M ≤ 100. ANSWER: between –6 and –5; between –3 and –2 If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. 64. To maximize the earnings, write a function that SOLUTION: relates the number and charge for each type of job. The sign of f (x) changes between the x values Since he charges $35 for each lawn mowed and between 0 and 1, and between 2 and 3. So the zeros $125 for each small landscaping job, F(M, L) = 35M of the quadratic function lies between 0 and 1, and + 125L. Next, substitute in the key values of M and between 2 and 3. L to determine the earnings for each combination of jobs. ANSWER: between 0 and 1; between 2 and 3 (M, L) F(M, L) = 35M + F(M, L) 125L

65. (100, 0) F(M, L) = 35(100) + 3500 125(0) SOLUTION: The sign of f (x) changes between the x values (55, 15) F(M, L) = 35(55) + 125 3800 between –1 and 0, and between 2 and 3. So the (15) zeros of the quadratic function lies between –1 and 0, and between 2 and 3. (0, 15) F(M, L) = 35(0) + 125 1875 (15) ANSWER: between –1 and 0; between 2 and 3 15 landscape jobs and 55 lawns; $3800 66. BUSINESS A landscaper can mow a lawn in 30 ANSWER: minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days 15 landscape jobs and 55 lawns; $3800 per week. He earns $35 per lawn and $125 per 2 landscape job. He cannot do more than 3 landscape Find each value if f (x) = 4x + 3, g(x) = –x , and jobs per day. Find the combination of lawns mowed h(x) = –2x2 – 2x + 4. and completed landscape jobs per week that will 67. f (–6) maximize income. Then find the maximum income. SOLUTION: SOLUTION: Substitute –6 for x in f (x). Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100. ANSWER: If the landscaper does 3 landscaping jobs per day, he –21 can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 68. g(–8) landscaping jobs, he can mow 55 lawns. SOLUTION: To maximize the earnings, write a function that Substitute –8 for x in g(x). relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of ANSWER: jobs. –64 (M, L) F(M, L) = 35M + F(M, L) 69. h(3) 125L SOLUTION: Substitute 3 for x in h(x). (100, 0) F(M, L) = 35(100) + 3500 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15) ANSWER: –20 15 landscape jobs and 55 lawns; $3800 70. f (c) ANSWER: 15 landscape jobs and 55 lawns; $3800 SOLUTION: Substitute c for x in f (x). Find each value if f (x) = 4x + 3, g(x) = –x2, and h(x) = –2x2 – 2x + 4. 67. f (–6) SOLUTION: ANSWER: Substitute –6 for x in f (x). 4c + 3

71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: –21

68. g(–8) ANSWER: 2 SOLUTION: –9d Substitute –8 for x in g(x). 72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –64 69. h(3) SOLUTION: Substitute 3 for x in h(x). ANSWER: –8b2 – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

Simplify.

SOLUTION: ANSWER:

2 ANSWER: 4. (2a – 4a – 8) ÷ (a + 1) 4y + 2x – 2 SOLUTION:

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) ANSWER:

SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

ANSWER: 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. 5 2 6. (y – 3y – 20) ÷ (y – 2) B. SOLUTION: C.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER: The correct choice is A. 4 3 2 y + 2y + 4y + 5y + 10 ANSWER: A 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) A. SOLUTION: B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

10.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2)

SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

10.

SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: ANSWER:

3a2b – 2ab2

11. 13.

SOLUTION: SOLUTION:

ANSWER: x + 3y –2

ANSWER: 14. 3y + 5 SOLUTION: Simplify

12.

SOLUTION:

ANSWER: ANSWER: 3a2b – 2ab2

15.

13. SOLUTION: SOLUTION:

ANSWER: ANSWER: 2 2a + b – 3 x + 3y –2

16. 14. SOLUTION: SOLUTION:

ANSWER: ANSWER: 4c2d2 – 6

15. 17.

SOLUTION: SOLUTION:

ANSWER: 3np – 6 + 7p ANSWER: 2 2a + b – 3 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 16. 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of SOLUTION: energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: ANSWER: 2 2 4c d – 6 –b + 8

19. BAKING The number of cookies produced in a 17. factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by SOLUTION: w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is:

ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs ANSWER: reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of Simplify. energy saved per CFL bulb. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: The average amount of energy saved per CFL bulb SOLUTION: is:

ANSWER: –b + 8 ANSWER: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced 21. (b3 – 4b2 + b – 2) ÷ (b + 1) per worker. SOLUTION: SOLUTION:

The average number of cookies produced per worker is:

ANSWER: ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

SOLUTION:

ANSWER: z3 – 2z2 – 4

ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

ANSWER: 24.

SOLUTION:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: z3 – 2z2 – 4 25. (g4 – 3g2 – 18) ÷ (g – 2) 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER: ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) 24. SOLUTION:

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: ANSWER:

27. ANSWER: SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

27. SOLUTION: SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

ANSWER:

Synthetic division:

28.

SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: ANSWER:

Synthetic division: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

ANSWER: Synthetic division:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

6 5 3 −1 SOLUTION: 31. (10y + 5y + 10y – 20y – 15)(5y + 5) Divide the function by the height (x + 2) to find the SOLUTION: length and width. Use synthetic division.

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 5-2 Dividing Polynomials ANSWER: 2x + 3, 3x + 5

32. CCSS REASONING A rectangular box for a new 33. PHYSICS The voltage V is related to current I and product is designed in such a way that the three dimensions always have a particular relationship power P by the equation . The power of a defined by the variable x. The volume of the box can 3 2 3 2 generator is modeled by P(t) = t + 9t + 26t + 24. If be written as 6x + 31x + 53x + 30, and the height is the current of the generator is I = t + 4, write an always x + 2. What are the width and length of the expression that represents the voltage. box? SOLUTION: SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

Synthetic division:

The depressed polynomial is .

ANSWER: 2 Since the volume of the box is the product of length, V(t) = t + 5t + 6 width and height, the width and length of box are (2x ENTERTAINMENT + 3)(3x + 5). 34. A magician gives these

instructions to a volunteer. • Choose a number and multiply it by 4. ANSWER: • Then add the sum of your number and 15 to the product you found. 2x + 3, 3x + 5 • Now divide by the sum of your number and 3. 33. PHYSICS The voltage V is related to current I and a. What number will the volunteer always have at the end? power P by the equation . The power of a b. Explain the process you used to discover the answer. generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an SOLUTION: expression that represents the voltage. a. Let x be the number.

SOLUTION:

Synthetic division: b. Sample answer: Let x be the number. Multiply the

x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the eSolutionsANSWER: Manual - Powered by Cognero product to get 5x + 15. Divide the polynomial by Pagex + 8 2 V(t) = t + 5t + 6 3. The quotient is 5.

34. ENTERTAINMENT A magician gives these 35. BUSINESS The number of magazine subscriptions instructions to a volunteer. sold can be estimated by ,where a is • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the the amount of money the company spent on product you found. advertising in hundreds of dollars and n is the number • Now divide by the sum of your number and 3. of subscriptions sold. a. What number will the volunteer always have at the end? a. Perform the division indicated by . b. Explain the process you used to discover the answer. b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: SOLUTION: a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is

the amount of money the company spent on Therefore, about 2423 subscriptions will be sold. advertising in hundreds of dollars and n is the number of subscriptions sold. ANSWER:

a. Perform the division indicated by . a. b. About how many subscriptions will be sold if b. about 2423 subscriptions $1500 is spent on advertising? Simplify. SOLUTION: 36. (x4 – y4) ÷ (x – y)

a. SOLUTION:

ANSWER: 2 2 b. There are 15 hundreds in 1500. Substitute a = 15. (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER:

Therefore, about 2423 subscriptions will be sold.

ANSWER: 3 2 2 –1 38. (a b – a b + 2b)(–ab) a. SOLUTION: b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION: ANSWER:

39.

ANSWER: SOLUTION: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER: n2 – n – 1

ANSWER: 40.

SOLUTION: 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER: ANSWER:

39. 41.

SOLUTION: SOLUTION:

ANSWER: ANSWER: 2 n – n – 1 3z 4 – z 3 + 2 z2 – 4z + 9–

40. 42. MULTIPLE REPRESENTATIONS Consider a 2 SOLUTION: rectangle with area 2x + 7x + 3 and length 2x + 1. a CONCRETE . Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? ANSWER: SOLUTION: a.

41.

SOLUTION:

The width is x + 3.

ANSWER: b.

3z 4 – z 3 + 2 z2 – 4z + 9–

c. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. Yes, the concrete model checks with the algebraic c. NUMERICAL Solve this problem algebraically model. using synthetic or long division. Does your concrete model check with your algebraic model? ANSWER: a. SOLUTION: a.

The width is x + 3.

The width is x + 3. 2 b. 2x + 7x + 3 ÷ (2x + 1)

c. c.

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the Yes, the concrete model checks with the algebraic remainder is 26. Jamal argues that the remainder is – model. 100. Is either of them correct? Explain your reasoning. ANSWER: a. SOLUTION: Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x The width is x + 3. + 3.

2 44. CHALLENGE If a polynomial is divided by a b. 2x + 7x + 3 ÷ (2x + 1) binomial and the remainder is 0, what does this tell

you about the relationship between the binomial and the polynomial? c. SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the yes binomial and the quotient.

43. ERROR ANALYSIS Sharon and Jamal are dividing ANSWER: 3 2 The binomial is a factor of the polynomial. 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 45. REASONING Review any of the division problems 100. Is either of them correct? Explain your in this lesson. What is the relationship between the reasoning. degrees of the dividend, the divisor, and the quotient? SOLUTION: SOLUTION: Use synthetic division. Sample answer: For example:

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell In this exercise, the degree of the dividend is 2. The you about the relationship between the binomial and degree of the divisor and the quotient are each 1. the polynomial? The degree of the quotient plus the degree of the SOLUTION: divisor equals the degree of the dividend. If a polynomial divided by a binomial has no ANSWER: remainder, then the polynomial has two factors: the Sample answer: The degree of the quotient plus the

binomial and the quotient. degree of the divisor equals the degree of the ANSWER: dividend. The binomial is a factor of the polynomial. 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. 45. REASONING Review any of the division problems in this lesson. What is the relationship between the SOLUTION: degrees of the dividend, the divisor, and the quotient? Sample answer: Begin by multiplying two binomials 2 SOLUTION: such as (x + 2)(x + 3) which simplifies to x + 5x + Sample answer: For example: 6. In order to get a remainder of 3 when divided, add 2 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two

polynomials is .

ANSWER:

Sample answer:

47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the divisor equals the degree of the dividend. ANSWER: SOLUTION: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the does not belong with the other three. The other

dividend. three expressions are polynomials. Since the 46. OPEN ENDED Write a quotient of two polynomials denominator of contains a variable, it is not a for which the remainder is 3. polynomial. SOLUTION: Sample answer: Begin by multiplying two binomials ANSWER: 2 such as (x + 2)(x + 3) which simplifies to x + 5x + does not belong with the other three. The other 6. In order to get a remainder of 3 when divided, add 2 3 to the trinomial to get x + 5x + 9. When divided, three expressions are polynomials. Since the there will be a remainder of 3. The quotient of two denominator of contains a variable, it is not a

polynomials is . polynomial.

ANSWER: 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction Sample answer: using the division of polynomials to make a paper cover for your textbook. 47. CCSS ARGUMENTS Identify the expression that SOLUTION: does not belong with the other three. Explain your Sample answer: By dividing the area of the paper reasoning. 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. SOLUTION: ANSWER: 2 does not belong with the other three. The other Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book three expressions are polynomials. Since the jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. denominator of contains a variable, it is not a polynomial. 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number ANSWER: of women?

does not belong with the other three. The other A

three expressions are polynomials. Since the B denominator of contains a variable, it is not a C polynomial.

D 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction using the division of polynomials to make a paper SOLUTION: cover for your textbook. Total number of employees: x + 3 Number of women: x SOLUTION: Ratio of the total number of employees to the number Sample answer: By dividing the area of the paper of women is: 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units The correct choice is A. long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is ANSWER: the width of each flap. A

ANSWER: 50. SAT/ACT Which polynomial has degree 3? 2 Sample answer: By dividing 140x + 60x by 10x, the A x3 + x2 –2x4 2 quotient of 14x + 6 provides the length of the book B –2x – 3x + 4 jacket. Then, subtracting 14x, we are left with 6 C 3x – 3 inches. Half of this length is the width of each flap. D x2 + x + 123 3 49. An office employs x women and 3 men. What is the E 1 + x + x ratio of the total number of employees to the number of women? SOLUTION: To determine the degree of a polynomial, look at the A term with the greatest exponent. The correct choice is E. B ANSWER:

C E

D 51. GRIDDED RESPONSE In the figure below, m + n + p = ? SOLUTION: Total number of employees: x + 3 Number of women: x Ratio of the total number of employees to the number of women is:

SOLUTION: The correct choice is A. Sum of the exterior angles of a triangle is . So, ANSWER: A ANSWER: 360 50. SAT/ACT Which polynomial has degree 3? 3 2 4 2 2 A x + x –2x 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2 2 B –2x – 3x + 4 F 2x C 3x – 3 H –10x2 + 17x 2 3 2 D x + x + 12 G –10x 3 E 1 + x + x J 2x2 + 17x SOLUTION: SOLUTION: To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is E.

ANSWER: The correct choice is H. E ANSWER: 51. GRIDDED RESPONSE In the figure below, H m + n + p = ? Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: Use the Distributive Property and then combine like terms. SOLUTION: Sum of the exterior angles of a triangle is . So,

ANSWER: 360 ANSWER: 2 2 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 3 2 2 3x + 2x + x + 4 F 2x 2 H –10x + 17x 3 2 2 54. (2y – 3y + 8) + (3y – 6y) G –10x 2 SOLUTION: J 2x + 17x SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

The correct choice is H. 55. 4a(2a – 3) + 3a(5a – 4)

ANSWER: SOLUTION: H Use the Distributive Property and then combine like terms. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: ANSWER: 2 Use the Distributive Property and then combine like 23a – 24a terms. 56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms.

ANSWER: ANSWER: 3 2 2 3 3x3 + 2x2 + x + 4 2c – 3c d – 2cd + 3d 2 2 3 3 2 57. (xy) (2xy z) 54. (2y – 3y + 8) + (3y – 6y) SOLUTION: SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 2y 3 + 3y 2 – 9y + 8 ANSWER: 55. 4a(2a – 3) + 3a(5a – 4) 8x5y 8z 3

SOLUTION: 2 –2 2 2 58. (3ab ) (2a b) Use the Distributive Property and then combine like terms. SOLUTION:

ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) ANSWER: SOLUTION: Use the FOIL method and then combine like terms.

59. LANDSCAPING Amado wants to plant a garden ANSWER: and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a 3 2 2 3 2c – 3c d – 2cd + 3d perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the 2 2 3 57. (xy) (2xy z) width of his garden be? SOLUTION: SOLUTION: Apply the properties of exponents to simplify the Let x be the length and y be the width of the garden. expression.

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) The solution of the inequality is SOLUTION: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft ANSWER: Solve each equation by completing the square.

60. x2 + 6x + 2 = 0 SOLUTION: 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 The solution of the inequality is SOLUTION: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. ANSWER: 60. x2 + 6x + 2 = 0

SOLUTION: 2 62. 2x + 6x + 5 = 0 SOLUTION:

ANSWER:

–3 ± ANSWER: 2 61. x – 8x – 3 = 0 SOLUTION: State the consecutive integers between which the zeros of each quadratic function are located.

63. SOLUTION: The sign of f (x) changes between the x values –6 ANSWER: and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

62. 2x2 + 6x + 5 = 0 ANSWER: between –6 and –5; between –3 and –2 SOLUTION:

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3 ANSWER:

65. SOLUTION: State the consecutive integers between which the zeros of each quadratic function are located. The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, 63. and between 2 and 3. SOLUTION: ANSWER: The sign of f (x) changes between the x values –6 between –1 and 0; between 2 and 3 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and 66. BUSINESS A landscaper can mow a lawn in 30 between –3 and –2. minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days ANSWER: per week. He earns $35 per lawn and $125 per between –6 and –5; between –3 and –2 landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. 64. SOLUTION: SOLUTION: Let M be the lawns mowed and L be the small The sign of f (x) changes between the x values landscaping jobs. Since the landscaper will do at most between 0 and 1, and between 2 and 3. So the zeros 3 landscaping jobs per day, L ≤ 15. Since he works of the quadratic function lies between 0 and 1, and 10 hours per day, 5 days a week, he can do at most between 2 and 3. 20 mowing jobs per day or 100 per week. So, M ≤ 100. ANSWER: between 0 and 1; between 2 and 3 If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. 65. SOLUTION: To maximize the earnings, write a function that relates the number and charge for each type of job. The sign of f (x) changes between the x values Since he charges $35 for each lawn mowed and between –1 and 0, and between 2 and 3. So the $125 for each small landscaping job, F(M, L) = 35M zeros of the quadratic function lies between –1 and 0, + 125L. Next, substitute in the key values of M and and between 2 and 3. L to determine the earnings for each combination of ANSWER: jobs.

between –1 and 0; between 2 and 3 (M, L) F(M, L) = 35M + F(M, L) 66. BUSINESS A landscaper can mow a lawn in 30 125L minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days (100, 0) F(M, L) = 35(100) + 3500 per week. He earns $35 per lawn and $125 per 125(0) landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed (55, 15) F(M, L) = 35(55) + 125 3800 and completed landscape jobs per week that will (15) maximize income. Then find the maximum income. SOLUTION: (0, 15) F(M, L) = 35(0) + 125 1875 (15) Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 15 landscape jobs and 55 lawns; $3800 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ ANSWER: 100. 15 landscape jobs and 55 lawns; $3800

If the landscaper does 3 landscaping jobs per day, he Find each value if f (x) = 4x + 3, g(x) = –x2, and can only do 11 mowing jobs per day and work a total h(x) = –2x2 – 2x + 4. of 10 hours per day. So, for the week, if he does 15 67. f (–6) landscaping jobs, he can mow 55 lawns. SOLUTION: To maximize the earnings, write a function that Substitute –6 for x in f (x). relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs. ANSWER: –21 (M, L) F(M, L) = 35M + F(M, L) 125L 68. g(–8) (100, 0) F(M, L) = 35(100) + 3500 SOLUTION: 125(0) Substitute –8 for x in g(x).

(55, 15) F(M, L) = 35(55) + 125 3800

(15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15) ANSWER: –64

15 landscape jobs and 55 lawns; $3800 69. h(3) ANSWER: SOLUTION: 15 landscape jobs and 55 lawns; $3800 Substitute 3 for x in h(x).

2 Find each value if f (x) = 4x + 3, g(x) = –x , and h(x) = –2x2 – 2x + 4. 67. f (–6) SOLUTION: Substitute –6 for x in f (x). ANSWER: –20 70. f (c) SOLUTION: ANSWER: Substitute c for x in f (x). –21

68. g(–8) SOLUTION: ANSWER: Substitute –8 for x in g(x). 4c + 3

71. g(3d) SOLUTION: Substitute 3d for x in g(x). ANSWER: –64 69. h(3) SOLUTION: ANSWER: Substitute 3 for x in h(x). 2 –9d 72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x). ANSWER: –8b2 – 12b

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

Simplify.

SOLUTION: ANSWER:

2 ANSWER: 4. (2a – 4a – 8) ÷ (a + 1) 4y + 2x – 2 SOLUTION:

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

2 3. (x – 6x – 20) ÷ (x + 2) ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10 ANSWER: 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. 5 2 6. (y – 3y – 20) ÷ (y – 2) B. SOLUTION: C.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER: The correct choice is A. 4 3 2 y + 2y + 4y + 5y + 10 ANSWER: 7. MULTIPLE CHOICE Which expression is equal A 2 –1 to (x + 3x – 9)(4 – x) ? Simplify. 2 A. 8. (10x + 15x + 20) ÷ (5x + 5) SOLUTION: B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

10.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

10.

SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: ANSWER:

3a2b – 2ab2

11. 13.

SOLUTION: SOLUTION:

ANSWER: x + 3y –2

ANSWER: 14. 3y + 5 SOLUTION: Simplify

12.

SOLUTION:

ANSWER: ANSWER:

3a2b – 2ab2

15. 13. SOLUTION: SOLUTION:

ANSWER: ANSWER: 2 x + 3y –2 2a + b – 3

14. 16.

SOLUTION: SOLUTION:

ANSWER: ANSWER:

4c2d2 – 6

15. 17.

SOLUTION: SOLUTION:

ANSWER: ANSWER: 3np – 6 + 7p 2 2a + b – 3 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 16. 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of SOLUTION: energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: ANSWER: 4c2d2 – 6 –b + 8

17. 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + SOLUTION: 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is:

ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste ANSWER: that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: The average amount of energy saved per CFL bulb SOLUTION: is:

ANSWER: –b + 8 ANSWER: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced 3 2 per worker. 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION: SOLUTION: The average number of cookies produced per worker is:

ANSWER: ANSWER:

4 3 2 –1 Simplify. 22. (z – 3z + 2z – 4z + 4)(z – 1) 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

SOLUTION:

ANSWER: z3 – 2z2 – 4 ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

ANSWER: 24.

SOLUTION: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: z3 – 2z2 – 4 4 2 5 3 2 25. (g – 3g – 18) ÷ (g – 2) 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER: ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) 24. SOLUTION: SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: ANSWER:

27. ANSWER: SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

27. SOLUTION: SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

ANSWER:

Synthetic division:

28.

SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6)

ANSWER: SOLUTION:

Synthetic division: 3 2 29. (2b – 6b + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division: ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1

SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: ANSWER:

2x + 3, 3x + 5

Synthetic division:

The depressed polynomial is .

ANSWER:

2 Since the volume of the box is the product of length, V(t) = t + 5t + 6 width and height, the width and length of box are (2x + 3)(3x + 5). 34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. ANSWER: • Then add the sum of your number and 15 to the 2x + 3, 3x + 5 product you found. • Now divide by the sum of your number and 3. 33. PHYSICS The voltage V is related to current I and a. What number will the volunteer always have at the end? power P by the equation . The power of a b. Explain the process you used to discover the answer. generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an SOLUTION: expression that represents the voltage. a. Let x be the number. SOLUTION:

Synthetic division: b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER:

a. 5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the ANSWER: 5-2 Dividing Polynomials product to get 5x + 15. Divide the polynomial by x + 2 V(t) = t + 5t + 6 3. The quotient is 5.

34. ENTERTAINMENT A magician gives these 35. BUSINESS The number of magazine subscriptions instructions to a volunteer. • Choose a number and multiply it by 4. sold can be estimated by ,where a is • Then add the sum of your number and 15 to the the amount of money the company spent on product you found. advertising in hundreds of dollars and n is the number • Now divide by the sum of your number and 3. of subscriptions sold. a. What number will the volunteer always have at the

end? a. Perform the division indicated by . b. Explain the process you used to discover the answer. b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: a. Let x be the number. SOLUTION:

b. Sample answer: Let x be the number. Multiply the

x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number Therefore, about 2423 subscriptions will be sold.

of subscriptions sold. ANSWER:

a. Perform the division indicated by . a. b. About how many subscriptions will be sold if b. about 2423 subscriptions $1500 is spent on advertising? Simplify. SOLUTION: 4 4 36. (x – y ) ÷ (x – y) a. SOLUTION:

eSolutions Manual - Powered by Cognero ANSWER: Page 9 2 2 b. There are 15 hundreds in 1500. Substitute a = 15. (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER:

Therefore, about 2423 subscriptions will be sold.

ANSWER: 3 2 2 –1 38. (a b – a b + 2b)(–ab) a. SOLUTION: b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y)

SOLUTION: ANSWER:

39.

ANSWER: SOLUTION: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER: n2 – n – 1

ANSWER: 40.

SOLUTION: 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER: ANSWER:

39. 41.

SOLUTION: SOLUTION:

ANSWER: ANSWER: n2 – n – 1

3z 4 – z 3 + 2 z2 – 4z + 9–

40. 42. MULTIPLE REPRESENTATIONS Consider a 2 SOLUTION: rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? ANSWER: SOLUTION: a.

41.

SOLUTION:

The width is x + 3.

ANSWER: b.

3z 4 – z 3 + 2 z2 – 4z + 9–

The width is x + 3. The width is x + 3. b. 2x 2 + 7x + 3 ÷ (2x + 1) b.

c. c.

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the Yes, the concrete model checks with the algebraic remainder is 26. Jamal argues that the remainder is – model. 100. Is either of them correct? Explain your ANSWER: reasoning. a. SOLUTION: Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x The width is x + 3. + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1) 44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? c. SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the yes binomial and the quotient.

43. ERROR ANALYSIS Sharon and Jamal are dividing ANSWER: 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the The binomial is a factor of the polynomial. remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your 45. REASONING Review any of the division problems reasoning. in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: SOLUTION: Use synthetic division. Sample answer: For example:

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

CHALLENGE 44. If a polynomial is divided by a binomial and the remainder is 0, what does this tell In this exercise, the degree of the dividend is 2. The you about the relationship between the binomial and degree of the divisor and the quotient are each 1. the polynomial? The degree of the quotient plus the degree of the SOLUTION: divisor equals the degree of the dividend. If a polynomial divided by a binomial has no ANSWER: remainder, then the polynomial has two factors: the binomial and the quotient. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the ANSWER: dividend. The binomial is a factor of the polynomial. 46. OPEN ENDED Write a quotient of two polynomials 45. REASONING Review any of the division problems for which the remainder is 3. in this lesson. What is the relationship between the SOLUTION: degrees of the dividend, the divisor, and the quotient? Sample answer: Begin by multiplying two binomials SOLUTION: such as (x + 2)(x + 3) which simplifies to x2 + 5x + Sample answer: For example: 6. In order to get a remainder of 3 when divided, add 2 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two

polynomials is .

ANSWER:

Sample answer:

47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

ANSWER: SOLUTION: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the does not belong with the other three. The other dividend. three expressions are polynomials. Since the OPEN ENDED 46. Write a quotient of two polynomials denominator of contains a variable, it is not a for which the remainder is 3. polynomial. SOLUTION: Sample answer: Begin by multiplying two binomials ANSWER: such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add does not belong with the other three. The other 2 3 to the trinomial to get x + 5x + 9. When divided, three expressions are polynomials. Since the there will be a remainder of 3. The quotient of two denominator of contains a variable, it is not a polynomials is . polynomial.

SOLUTION: ANSWER: 2 does not belong with the other three. The other Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book three expressions are polynomials. Since the jacket. Then, subtracting 14x, we are left with 6 denominator of contains a variable, it is not a inches. Half of this length is the width of each flap. polynomial. 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number ANSWER: of women?

does not belong with the other three. The other A three expressions are polynomials. Since the B denominator of contains a variable, it is not a

C polynomial.

48. WRITING IN MATH Use the information at the D beginning of the lesson to write assembly instruction using the division of polynomials to make a paper SOLUTION: cover for your textbook. Total number of employees: x + 3 SOLUTION: Number of women: x Ratio of the total number of employees to the number Sample answer: By dividing the area of the paper 2 of women is: 140x + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units The correct choice is A. long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is ANSWER:

the width of each flap. A ANSWER: 50. SAT/ACT Which polynomial has degree 3? 2 Sample answer: By dividing 140x + 60x by 10x, the A x3 + x2 –2x4 quotient of 14x + 6 provides the length of the book 2 jacket. Then, subtracting 14x, we are left with 6 B –2x – 3x + 4 inches. Half of this length is the width of each flap. C 3x – 3 D x2 + x + 123 49. An office employs x women and 3 men. What is the E 3 ratio of the total number of employees to the number 1 + x + x of women? SOLUTION:

A To determine the degree of a polynomial, look at the term with the greatest exponent.

B The correct choice is E. ANSWER: C E

D 51. GRIDDED RESPONSE In the figure below, m + n + p = ? SOLUTION: Total number of employees: x + 3 Number of women: x Ratio of the total number of employees to the number of women is:

SOLUTION: The correct choice is A. Sum of the exterior angles of a triangle is . So, ANSWER: A ANSWER: 360 50. SAT/ACT Which polynomial has degree 3? 3 2 4 A x + x –2x 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 2 B –2x – 3x + 4 F 2x C 3x – 3 H –10x2 + 17x 2 3 2 D x + x + 12 G –10x 3 E 1 + x + x J 2x2 + 17x SOLUTION: SOLUTION: To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is E.

terms. SOLUTION: Sum of the exterior angles of a triangle is . So,

ANSWER: 360 2 2 ANSWER: 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 3 2 2 3x + 2x + x + 4 F 2x H 2 –10x + 17x 3 2 2 54. (2y – 3y + 8) + (3y – 6y) G –10x J 2x2 + 17x SOLUTION: SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

The correct choice is H. 55. 4a(2a – 3) + 3a(5a – 4) ANSWER: SOLUTION: H Use the Distributive Property and then combine like terms. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: ANSWER: Use the Distributive Property and then combine like 2 23a – 24a terms. 56. (c + d)(c – d)(2c – 3d)

SOLUTION: Use the FOIL method and then combine like terms.

ANSWER: ANSWER: 3 2 3x + 2x + x + 4 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 3 2 54. (2y – 3y + 8) + (3y – 6y) 57. (xy) (2xy z) SOLUTION: SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 2y 3 + 3y 2 – 9y + 8 ANSWER: 55. 4a(2a – 3) + 3a(5a – 4) 8x5y 8z 3 SOLUTION: 2 –2 2 2 Use the Distributive Property and then combine like 58. (3ab ) (2a b)

terms. SOLUTION:

ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) ANSWER: SOLUTION:

Use the FOIL method and then combine like terms.

59. LANDSCAPING Amado wants to plant a garden ANSWER: and surround it with decorative stones. He has 3 2 2 3 enough stones to enclose a rectangular garden with a 2c – 3c d – 2cd + 3d perimeter of 68 feet, but he wants the garden to 2 2 3 cover no more than 240 square feet. What could the 57. (xy) (2xy z) width of his garden be? SOLUTION: SOLUTION: Apply the properties of exponents to simplify the Let x be the length and y be the width of the garden. expression.

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION: The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft ANSWER: Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION: 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 The solution of the inequality is SOLUTION: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. ANSWER: 60. x2 + 6x + 2 = 0 SOLUTION: 62. 2x2 + 6x + 5 = 0 SOLUTION:

ANSWER:

–3 ± ANSWER: 2 61. x – 8x – 3 = 0 SOLUTION: State the consecutive integers between which the zeros of each quadratic function are located.

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3 ANSWER:

65.

State the consecutive integers between which SOLUTION: the zeros of each quadratic function are located. The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, 63. and between 2 and 3.

SOLUTION: ANSWER: The sign of f (x) changes between the x values –6 between –1 and 0; between 2 and 3 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and 66. BUSINESS A landscaper can mow a lawn in 30 between –3 and –2. minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days ANSWER: per week. He earns $35 per lawn and $125 per between –6 and –5; between –3 and –2 landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. 64. SOLUTION: SOLUTION: Let M be the lawns mowed and L be the small The sign of f (x) changes between the x values landscaping jobs. Since the landscaper will do at most between 0 and 1, and between 2 and 3. So the zeros 3 landscaping jobs per day, L ≤ 15. Since he works of the quadratic function lies between 0 and 1, and 10 hours per day, 5 days a week, he can do at most between 2 and 3. 20 mowing jobs per day or 100 per week. So, M ≤ 100. ANSWER:

between 0 and 1; between 2 and 3 If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. 65.

SOLUTION: To maximize the earnings, write a function that The sign of f (x) changes between the x values relates the number and charge for each type of job. between –1 and 0, and between 2 and 3. So the Since he charges $35 for each lawn mowed and zeros of the quadratic function lies between –1 and 0, $125 for each small landscaping job, F(M, L) = 35M and between 2 and 3. + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of ANSWER: jobs. between –1 and 0; between 2 and 3 (M, L) F(M, L) = 35M + F(M, L) 66. BUSINESS A landscaper can mow a lawn in 30 125L minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days (100, 0) F(M, L) = 35(100) + 3500 per week. He earns $35 per lawn and $125 per 125(0) landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed (55, 15) F(M, L) = 35(55) + 125 3800 and completed landscape jobs per week that will (15) maximize income. Then find the maximum income. SOLUTION: (0, 15) F(M, L) = 35(0) + 125 1875 Let M be the lawns mowed and L be the small (15) landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 15 landscape jobs and 55 lawns; $3800 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ ANSWER: 100. 15 landscape jobs and 55 lawns; $3800

If the landscaper does 3 landscaping jobs per day, he Find each value if f (x) = 4x + 3, g(x) = –x2, and can only do 11 mowing jobs per day and work a total 2 of 10 hours per day. So, for the week, if he does 15 h(x) = –2x – 2x + 4. landscaping jobs, he can mow 55 lawns. 67. f (–6) SOLUTION: To maximize the earnings, write a function that Substitute –6 for x in f (x). relates the number and charge for each type of job.

Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs. ANSWER: –21 (M, L) F(M, L) = 35M + F(M, L) 125L 68. g(–8) (100, 0) F(M, L) = 35(100) + 3500 SOLUTION: 125(0) Substitute –8 for x in g(x). (55, 15) F(M, L) = 35(55) + 125 3800 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15) ANSWER: –64 15 landscape jobs and 55 lawns; $3800 69. h(3) ANSWER: SOLUTION: 15 landscape jobs and 55 lawns; $3800 Substitute 3 for x in h(x). 2 Find each value if f (x) = 4x + 3, g(x) = –x , and h(x) = –2x2 – 2x + 4. 67. f (–6) SOLUTION: Substitute –6 for x in f (x). ANSWER: –20 70. f (c) SOLUTION: ANSWER: Substitute c for x in f (x). –21

68. g(–8)

SOLUTION: ANSWER: Substitute –8 for x in g(x). 4c + 3

71. g(3d) SOLUTION: ANSWER: Substitute 3d for x in g(x).

–64

69. h(3) SOLUTION: ANSWER: Substitute 3 for x in h(x). 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x). ANSWER:

–8b2 – 12b

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

1. ANSWER:

SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

2 3. (x – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? SOLUTION: A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION: The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? 2 9. (18a + 6a + 9) ÷ (3a – 2) A. SOLUTION: B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

10.

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) ANSWER: SOLUTION: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: 3a2b – 2ab2 ANSWER:

13.

10. SOLUTION:

SOLUTION:

ANSWER: x + 3y –2

14. ANSWER: SOLUTION:

11.

SOLUTION: ANSWER:

15.

SOLUTION:

ANSWER: 3y + 5 Simplify

12. ANSWER: 2 2a + b – 3 SOLUTION:

16.

ANSWER: SOLUTION: 3a2b – 2ab2

13.

SOLUTION: ANSWER: 4c2d2 – 6

17.

SOLUTION: ANSWER: x + 3y –2

14.

SOLUTION: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of ANSWER: bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb 15. is:

SOLUTION:

ANSWER: –b + 8

SOLUTION: 16. The average number of cookies produced per worker is: SOLUTION:

ANSWER:

ANSWER: Simplify. 2 2 2 4c d – 6 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: 17.

SOLUTION:

ANSWER:

ANSWER: 3np – 6 + 7p 21. (b3 – 4b2 + b – 2) ÷ (b + 1) 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste SOLUTION: that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is: ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) ANSWER: –b + 8 SOLUTION:

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. ANSWER: SOLUTION: z3 – 2z2 – 4 The average number of cookies produced per worker 5 3 2 is: 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

ANSWER:

ANSWER: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

24.

SOLUTION:

ANSWER:

ANSWER: 3 2 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

ANSWER: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

24.

SOLUTION: 27.

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER: ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) 28. SOLUTION:

SOLUTION:

ANSWER:

ANSWER: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

27.

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6)

SOLUTION:

ANSWER: Synthetic division:

28.

SOLUTION:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 Divide the function by the height (x + 2) to find the 30. (6z + 3z – 9z )(3z – 6) length and width. Use synthetic division. SOLUTION:

Synthetic division:

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x

ANSWER: + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 33. PHYSICS The voltage V is related to current I and SOLUTION: power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

Synthetic division:

ANSWER:

32. CCSS REASONING A rectangular box for a new ANSWER: 2 product is designed in such a way that the three V(t) = t + 5t + 6 dimensions always have a particular relationship defined by the variable x. The volume of the box can 34. ENTERTAINMENT A magician gives these 3 2 be written as 6x + 31x + 53x + 30, and the height is instructions to a volunteer. always x + 2. What are the width and length of the • Choose a number and multiply it by 4. box? • Then add the sum of your number and 15 to the product you found. SOLUTION: • Now divide by the sum of your number and 3. Divide the function by the height (x + 2) to find the a. What number will the volunteer always have at the length and width. Use synthetic division. end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

The depressed polynomial is .

b. Sample answer: Let x be the number. Multiply the Since the volume of the box is the product of length, x by 4 to get 4x. Then add x +15 to the product to get width and height, the width and length of box are (2x 5x + 15. Divide the polynomial by x + 3. The quotient + 3)(3x + 5). is 5. ANSWER: ANSWER: a.5 2x + 3, 3x + 5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the 33. PHYSICS The voltage V is related to current I and product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. power P by the equation . The power of a 35. BUSINESS The number of magazine subscriptions generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an sold can be estimated by ,where a is expression that represents the voltage. the amount of money the company spent on SOLUTION: advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

Synthetic division: b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION:

ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer. b. There are 15 hundreds in 1500. Substitute a = 15. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number. Therefore, about 2423 subscriptions will be sold.

ANSWER:

a. b. about 2423 subscriptions b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get Simplify. 5x + 15. Divide the polynomial by x + 3. The quotient 4 4 is 5. 36. (x – y ) ÷ (x – y) SOLUTION: ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions ANSWER: 2 2 (x + y )(x + y) sold can be estimated by ,where a is the amount of money the company spent on 37. (28c3d2 – 21cd2) ÷ (14cd) advertising in hundreds of dollars and n is the number of subscriptions sold. SOLUTION:

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? ANSWER: SOLUTION:

a. 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

39.

SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER:

a. 5-2 Dividing Polynomials ANSWER: b. about 2423 subscriptions n2 – n – 1

Simplify. 36. (x4 – y4) ÷ (x – y) 40. SOLUTION: SOLUTION:

ANSWER: 2 2 (x + y )(x + y) ANSWER:

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: 41.

SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab)

SOLUTION: ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 ANSWER: rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. 39. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete SOLUTION: model check with your algebraic model? SOLUTION: a.

ANSWER: n2 – n – 1 eSolutions Manual - Powered by Cognero Page 10 40.

SOLUTION: The width is x + 3.

c. ANSWER:

41. Yes, the concrete model checks with the algebraic model.

SOLUTION: ANSWER: a.

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. The width is x + 3. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. 2x 2 + 7x + 3 ÷ (2x + 1) b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete c. model check with your algebraic model? SOLUTION: a. yes 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

The width is x + 3.

b. The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

c. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and Yes, the concrete model checks with the algebraic the polynomial? model. SOLUTION: ANSWER: If a polynomial divided by a binomial has no a. remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example:

The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 In this exercise, the degree of the dividend is 2. The 2x – 4x + 3x – 1 by x – 3. Sharon claims that the degree of the divisor and the quotient are each 1. remainder is 26. Jamal argues that the remainder is – The degree of the quotient plus the degree of the 100. Is either of them correct? Explain your divisor equals the degree of the dividend. reasoning. SOLUTION: ANSWER: Use synthetic division. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. SOLUTION: The remainder is 26. So, Sharon is correct. Jamal Sample answer: Begin by multiplying two binomials actually divided by x + 3. such as (x + 2)(x + 3) which simplifies to x2 + 5x + ANSWER: 6. In order to get a remainder of 3 when divided, add 2 Sample answer: Sharon; Jamal actually divided by x 3 to the trinomial to get x + 5x + 9. When divided, + 3. there will be a remainder of 3. The quotient of two

44. CHALLENGE If a polynomial is divided by a polynomials is . binomial and the remainder is 0, what does this tell you about the relationship between the binomial and ANSWER: the polynomial? Sample answer: SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the 47. CCSS ARGUMENTS Identify the expression that binomial and the quotient. does not belong with the other three. Explain your reasoning. ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: SOLUTION: Sample answer: For example: does not belong with the other three. The other

three expressions are polynomials. Since the

denominator of contains a variable, it is not a polynomial.

ANSWER:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a

polynomial. In this exercise, the degree of the dividend is 2. The WRITING IN MATH degree of the divisor and the quotient are each 1. 48. Use the information at the The degree of the quotient plus the degree of the beginning of the lesson to write assembly instruction divisor equals the degree of the dividend. using the division of polynomials to make a paper cover for your textbook. ANSWER: SOLUTION: Sample answer: The degree of the quotient plus the Sample answer: By dividing the area of the paper degree of the divisor equals the degree of the 2 dividend. 140x + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book 46. OPEN ENDED Write a quotient of two polynomials jacket. The front and back cover are each 6x units for which the remainder is 3. long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is SOLUTION: the width of each flap. Sample answer: Begin by multiplying two binomials ANSWER: such as (x + 2)(x + 3) which simplifies to x2 + 5x + 2 6. In order to get a remainder of 3 when divided, add Sample answer: By dividing 140x + 60x by 10x, the 2 3 to the trinomial to get x + 5x + 9. When divided, quotient of 14x + 6 provides the length of the book there will be a remainder of 3. The quotient of two jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. polynomials is . 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number ANSWER: of women?

Sample answer: A

47. CCSS ARGUMENTS Identify the expression that B does not belong with the other three. Explain your reasoning. C

SOLUTION: Total number of employees: x + 3 SOLUTION: Number of women: x Ratio of the total number of employees to the number does not belong with the other three. The other of women is:

three expressions are polynomials. Since the denominator of contains a variable, it is not a The correct choice is A. polynomial. ANSWER: A ANSWER: 50. SAT/ACT Which polynomial has degree 3? does not belong with the other three. The other A x3 + x2 –2x4 2 three expressions are polynomials. Since the B –2x – 3x + 4 denominator of contains a variable, it is not a C 3x – 3 D x2 + x + 123 polynomial. 3 E 1 + x + x 48. WRITING IN MATH Use the information at the SOLUTION: beginning of the lesson to write assembly instruction To determine the degree of a polynomial, look at the using the division of polynomials to make a paper cover for your textbook. term with the greatest exponent. The correct choice is E. SOLUTION: Sample answer: By dividing the area of the paper ANSWER: 140x2 + 60x by the height of the book jacket 10x, the E quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units 51. GRIDDED RESPONSE In the figure below, long and the spine is 2x units long. Then, subtracting m + n + p = ? 14x, we are left with 6 inches. Half of this length is the width of each flap.

ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 SOLUTION: inches. Half of this length is the width of each flap. Sum of the exterior angles of a triangle is . So, 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number ANSWER:

of women? 360

A 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 B F 2x H –10x2 + 17x 2 C G –10x 2 J 2x + 17x D SOLUTION: SOLUTION: Total number of employees: x + 3 Number of women: x Ratio of the total number of employees to the number of women is: The correct choice is H.

ANSWER: The correct choice is A. H

ANSWER: Simplify. 3 2 3 A 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: 50. SAT/ACT Which polynomial has degree 3? Use the Distributive Property and then combine like A 3 2 4 x + x –2x terms. 2 B –2x – 3x + 4 C 3x – 3 D x2 + x + 123 3 E 1 + x + x SOLUTION: To determine the degree of a polynomial, look at the ANSWER:

term with the greatest exponent. 3 2 The correct choice is E. 3x + 2x + x + 4

ANSWER: 3 2 54. (2y – 3y + 8) + (3y – 6y) E SOLUTION: 51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: SOLUTION: Sum of the exterior angles of a triangle is . Use the Distributive Property and then combine like terms. So,

ANSWER: 360 ANSWER: 2 2 2 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 23a – 24a 2 F 2x 56. (c + d)(c – d)(2c – 3d) H –10x2 + 17x SOLUTION: G 2 –10x Use the FOIL method and then combine like terms. 2 J 2x + 17x SOLUTION: ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) The correct choice is H. SOLUTION: ANSWER: Apply the properties of exponents to simplify the H expression. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x)

SOLUTION: ANSWER: Use the Distributive Property and then combine like 8x5y 8z 3 terms. 2 –2 2 2 58. (3ab ) (2a b)

SOLUTION:

ANSWER: 3x3 + 2x2 + x + 4

ANSWER: 3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has ANSWER: enough stones to enclose a rectangular garden with a 2y 3 + 3y 2 – 9y + 8 perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the 55. 4a(2a – 3) + 3a(5a – 4) width of his garden be? SOLUTION: SOLUTION: Use the Distributive Property and then combine like Let x be the length and y be the width of the garden. terms.

ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: The solution of the inequality is Use the FOIL method and then combine like terms. . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3 ANSWER: 0 to 10 ft or 24 to 34 ft 2 2 3 57. (xy) (2xy z) Solve each equation by completing the square. SOLUTION: 60. x2 + 6x + 2 = 0 Apply the properties of exponents to simplify the expression. SOLUTION:

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 ANSWER: SOLUTION:

59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? ANSWER: SOLUTION: Let x be the length and y be the width of the garden. 62. 2x2 + 6x + 5 = 0

SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 2 60. x + 6x + 2 = 0 State the consecutive integers between which the zeros of each quadratic function are located. SOLUTION:

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

ANSWER:

–3 ± 64. SOLUTION: 2 61. x – 8x – 3 = 0 The sign of f (x) changes between the x values SOLUTION: between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3

65. ANSWER: SOLUTION: The sign of f (x) changes between the x values 62. 2x2 + 6x + 5 = 0 between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, SOLUTION: and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

ANSWER: SOLUTION: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most State the consecutive integers between which 20 mowing jobs per day or 100 per week. So, M the zeros of each quadratic function are located. ≤ 100.

63. If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total SOLUTION: of 10 hours per day. So, for the week, if he does 15 The sign of f (x) changes between the x values –6 landscaping jobs, he can mow 55 lawns. and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and To maximize the earnings, write a function that between –3 and –2. relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and ANSWER: $125 for each small landscaping job, F(M, L) = 35M between –6 and –5; between –3 and –2 + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

64. (M, L) F(M, L) = 35M + F(M, L) SOLUTION: 125L The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros (100, 0) F(M, L) = 35(100) + 3500 of the quadratic function lies between 0 and 1, and 125(0) between 2 and 3. (55, 15) F(M, L) = 35(55) + 125 3800 ANSWER: (15) between 0 and 1; between 2 and 3 (0, 15) F(M, L) = 35(0) + 125 1875 (15)

65. 15 landscape jobs and 55 lawns; $3800 SOLUTION: The sign of f (x) changes between the x values ANSWER: between –1 and 0, and between 2 and 3. So the 15 landscape jobs and 55 lawns; $3800 zeros of the quadratic function lies between –1 and 0, and between 2 and 3. Find each value if f (x) = 4x + 3, g(x) = –x2, and h(x) = –2x2 – 2x + 4. ANSWER: 67. f (–6) between –1 and 0; between 2 and 3 SOLUTION: 66. BUSINESS A landscaper can mow a lawn in 30 Substitute –6 for x in f (x). minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. ANSWER: –21 SOLUTION: Let M be the lawns mowed and L be the small 68. g(–8) landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works SOLUTION: 10 hours per day, 5 days a week, he can do at most Substitute –8 for x in g(x). 20 mowing jobs per day or 100 per week. So, M ≤ 100.

To maximize the earnings, write a function that 69. h(3) relates the number and charge for each type of job. SOLUTION: Since he charges $35 for each lawn mowed and Substitute 3 for x in h(x). $125 for each small landscaping job, F(M, L) = 35M

+ 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) 125L ANSWER: (100, 0) F(M, L) = 35(100) + 3500 –20 125(0) 70. f (c) (55, 15) F(M, L) = 35(55) + 125 3800 SOLUTION: (15) Substitute c for x in f (x). (0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800 ANSWER: ANSWER: 4c + 3 15 landscape jobs and 55 lawns; $3800 71. g(3d) Find each value if f (x) = 4x + 3, g(x) = –x2, and SOLUTION: h(x) = –2x2 – 2x + 4. Substitute 3d for x in g(x). 67. f (–6)

SOLUTION: Substitute –6 for x in f (x). ANSWER: 2 –9d

72. h(2b + 1) ANSWER: SOLUTION: –21 Substitute 2b + 1 for x in h(x).

68. g(–8) SOLUTION: Substitute –8 for x in g(x).

ANSWER: –8b2 – 12b ANSWER: –64 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

1. ANSWER:

SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

2 3. (x – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? SOLUTION: A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION: The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? 2 9. (18a + 6a + 9) ÷ (3a – 2) A. SOLUTION: B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

10.

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) ANSWER: SOLUTION: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: 3a2b – 2ab2 ANSWER:

13.

10. SOLUTION:

SOLUTION:

ANSWER: x + 3y –2

14. ANSWER: SOLUTION:

11.

SOLUTION: ANSWER:

15.

SOLUTION:

ANSWER: 3y + 5 Simplify

12. ANSWER: 2 2a + b – 3 SOLUTION:

16.

ANSWER: SOLUTION: 3a2b – 2ab2

13.

SOLUTION: ANSWER: 4c2d2 – 6

17.

SOLUTION: ANSWER: x + 3y –2

14.

SOLUTION: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of ANSWER: bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb 15. is:

SOLUTION:

ANSWER: –b + 8

SOLUTION: 16. The average number of cookies produced per worker is: SOLUTION:

ANSWER:

ANSWER: Simplify. 2 2 2 4c d – 6 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: 17.

SOLUTION:

ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) ANSWER: –b + 8 SOLUTION:

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. ANSWER: SOLUTION: z3 – 2z2 – 4 The average number of cookies produced per worker 5 3 2 is: 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

ANSWER:

ANSWER: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

24.

SOLUTION:

ANSWER:

ANSWER: 3 2 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

ANSWER: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

24.

SOLUTION: 27.

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER: ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) 28. SOLUTION:

SOLUTION:

ANSWER:

ANSWER: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

27.

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6)

SOLUTION:

ANSWER: Synthetic division:

28.

SOLUTION:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 Divide the function by the height (x + 2) to find the 30. (6z + 3z – 9z )(3z – 6) length and width. Use synthetic division. SOLUTION:

Synthetic division:

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x

ANSWER: + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 33. PHYSICS The voltage V is related to current I and SOLUTION: power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

Synthetic division:

ANSWER:

The depressed polynomial is .

a. Perform the division indicated by .

Synthetic division: b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION:

ANSWER: 2 V(t) = t + 5t + 6

ANSWER:

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? ANSWER: SOLUTION:

a. 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

39.

SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER:

a. ANSWER: b. about 2423 subscriptions n2 – n – 1

Simplify. 36. (x4 – y4) ÷ (x – y) 40. SOLUTION: SOLUTION:

ANSWER: 2 2 (x + y )(x + y) ANSWER:

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: 41.

SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab)

SOLUTION: ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

ANSWER: n2 – n – 1

40.

SOLUTION: The width is x + 3.

c. ANSWER:

41. Yes, the concrete model checks with the algebraic model.

SOLUTION: ANSWER: a.

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. The width is x + 3. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. 2x 2 + 7x + 3 ÷ (2x + 1) b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete c. model check with your algebraic model? 5-2 Dividing Polynomials SOLUTION: a. yes 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

The width is x + 3.

b. The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

c. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example:

The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes eSolutions Manual - Powered by Cognero Page 11 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 In this exercise, the degree of the dividend is 2. The 2x – 4x + 3x – 1 by x – 3. Sharon claims that the degree of the divisor and the quotient are each 1. remainder is 26. Jamal argues that the remainder is – The degree of the quotient plus the degree of the 100. Is either of them correct? Explain your divisor equals the degree of the dividend. reasoning. SOLUTION: ANSWER: Use synthetic division. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

three expressions are polynomials. Since the

denominator of contains a variable, it is not a polynomial.

ANSWER:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a

Sample answer: A

47. CCSS ARGUMENTS Identify the expression that B does not belong with the other three. Explain your reasoning. C

SOLUTION: Total number of employees: x + 3 SOLUTION: Number of women: x Ratio of the total number of employees to the number does not belong with the other three. The other of women is:

of women? 360

ANSWER: The correct choice is A. H

term with the greatest exponent. 3 2 The correct choice is E. 3x + 2x + x + 4

ANSWER: 3 2 54. (2y – 3y + 8) + (3y – 6y) E SOLUTION: 51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: SOLUTION: Sum of the exterior angles of a triangle is . Use the Distributive Property and then combine like terms. So,

2 2 3 57. (xy) (2xy z) The correct choice is H. SOLUTION: ANSWER: Apply the properties of exponents to simplify the H expression. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x)

SOLUTION: ANSWER: Use the Distributive Property and then combine like 8x5y 8z 3 terms. 2 –2 2 2 58. (3ab ) (2a b)

SOLUTION:

ANSWER: 3x3 + 2x2 + x + 4

ANSWER: 3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

ANSWER: 2 23a – 24a

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 ANSWER: SOLUTION:

SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

ANSWER:

ANSWER: between 0 and 1; between 2 and 3

ANSWER: between –1 and 0; between 2 and 3

+ 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

SOLUTION: Substitute –6 for x in f (x). ANSWER: 2 –9d

72. h(2b + 1) ANSWER: SOLUTION: –21 Substitute 2b + 1 for x in h(x).

68. g(–8) SOLUTION: Substitute –8 for x in g(x).

ANSWER: –8b2 – 12b ANSWER: –64 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

SOLUTION:

ANSWER:

ANSWER: 4y + 2x – 2 5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) 2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION: SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

ANSWER: 5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION: 2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION: ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. ANSWER:

C. 2 9. (18a + 6a + 9) ÷ (3a – 2) D. SOLUTION: SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

10. The correct choice is A.

ANSWER: SOLUTION: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

11.

SOLUTION: ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: 3y + 5 Simplify

12.

ANSWER: SOLUTION:

ANSWER: 10. 3a2b – 2ab2 SOLUTION:

13.

SOLUTION:

ANSWER: ANSWER:

x + 3y –2

14. 11.

SOLUTION: SOLUTION:

ANSWER:

15. ANSWER: 3y + 5 SOLUTION: Simplify

12.

SOLUTION:

ANSWER: ANSWER: 2 2a + b – 3 3a2b – 2ab2

16. 13. SOLUTION: SOLUTION:

ANSWER: ANSWER: 4c2d2 – 6 x + 3y –2

17. 14. SOLUTION: SOLUTION:

ANSWER: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 15. 2 be estimated by –b + 8b, where b is the number of SOLUTION: bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: ANSWER: 2 2a + b – 3 –b + 8

19. BAKING The number of cookies produced in a 16. factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by SOLUTION: w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is:

ANSWER: 4c2d2 – 6 ANSWER:

17.

SOLUTION: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs ANSWER: reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. 21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: ANSWER: –b + 8

19. BAKING The number of cookies produced in a 2 factory each day can be estimated by –w + 16w + 4 3 2 –1 1000, where w is the number of workers. Divide by 22. (z – 3z + 2z – 4z + 4)(z – 1) w to find the average number of cookies produced SOLUTION: per worker.

SOLUTION: The average number of cookies produced per worker is:

ANSWER: z3 – 2z2 – 4

ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER:

24. ANSWER:

SOLUTION:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

ANSWER: 25. (g4 – 3g2 – 18) ÷ (g – 2)

SOLUTION:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: 2 z3 – 2z2 – 4 26. (6a – 3a + 9) ÷ (3a – 2)

5 3 2 SOLUTION: 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION:

ANSWER:

24.

SOLUTION: ANSWER:

27.

SOLUTION: ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION: ANSWER:

28.

SOLUTION:

ANSWER:

27. 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

28.

SOLUTION: Synthetic division:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

ANSWER: CCSS REASONING 32. A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship 6 4 2 –1 defined by the variable x. The volume of the box can 30. (6z + 3z – 9z )(3z – 6) 3 2 be written as 6x + 31x + 53x + 30, and the height is SOLUTION: always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the Synthetic division: length and width. Use synthetic division.

The depressed polynomial is .

ANSWER:

6 5 3 −1 Since the volume of the box is the product of length, 31. (10y + 5y + 10y – 20y – 15)(5y + 5) width and height, the width and length of box are (2x

SOLUTION: + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

ANSWER: Synthetic division:

box? 2 V(t) = t + 5t + 6 SOLUTION: Divide the function by the height (x + 2) to find the 34. ENTERTAINMENT A magician gives these length and width. Use synthetic division. instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? The depressed polynomial is . b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: b. Sample answer: Let x be the number. Multiply the 2x + 3, 3x + 5 x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient 33. PHYSICS The voltage V is related to current I and is 5. power P by the equation . The power of a ANSWER: 3 2 a.5 generator is modeled by P(t) = t + 9t + 26t + 24. If b. Sample answer: Let x be the number. the current of the generator is I = t + 4, write an Multiply the x by 4 to get 4x. Then add x + 15 to the expression that represents the voltage. product to get 5x + 15. Divide the polynomial by x + SOLUTION: 3. The quotient is 5. 35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is Synthetic division: the amount of money the company spent on

advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: ANSWER:

2 V(t) = t + 5t + 6 a. 34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. There are 15 hundreds in 1500. Substitute a = 15. b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the Therefore, about 2423 subscriptions will be sold. x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient ANSWER: is 5. a. ANSWER: a.5 b. about 2423 subscriptions b. Sample answer: Let x be the number. Simplify. Multiply the x by 4 to get 4x. Then add x + 15 to the 4 4 product to get 5x + 15. Divide the polynomial by x + 36. (x – y ) ÷ (x – y) 3. The quotient is 5. SOLUTION: 35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number

of subscriptions sold. ANSWER: 2 2 a. Perform the division indicated by . (x + y )(x + y) b. About how many subscriptions will be sold if $1500 is spent on advertising? 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) b. There are 15 hundreds in 1500. Substitute a = 15. SOLUTION:

ANSWER:

Therefore, about 2423 subscriptions will be sold. 39.

ANSWER: SOLUTION:

a. b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION: ANSWER: n2 – n – 1

40.

SOLUTION: ANSWER: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: ANSWER:

ANSWER:

41.

SOLUTION: 3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER: ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

39. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. SOLUTION: a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: ANSWER: a. n2 – n – 1

40.

SOLUTION:

The width is x + 3. ANSWER: b.

c. 41.

SOLUTION:

Yes, the concrete model checks with the algebraic model.

ANSWER: a. ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete The width is x + 3. model check with your algebraic model? b. 2x 2 + 7x + 3 ÷ (2x + 1) SOLUTION:

yes

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Yes, the concrete model checks with the algebraic Sample answer: Sharon; Jamal actually divided by x model. + 3. ANSWER: 44. CHALLENGE If a polynomial is divided by a a. binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems The width is x + 3. in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? 2 b. 2x + 7x + 3 ÷ (2x + 1) SOLUTION: Sample answer: For example:

yes

ANSWER: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the The remainder is 26. So, Sharon is correct. Jamal dividend. actually divided by x + 3.

ANSWER: 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. Sample answer: Sharon; Jamal actually divided by x + 3. SOLUTION: Sample answer: Begin by multiplying two binomials 44. CHALLENGE If a polynomial is divided by a 2 binomial and the remainder is 0, what does this tell such as (x + 2)(x + 3) which simplifies to x + 5x + you about the relationship between the binomial and 6. In order to get a remainder of 3 when divided, add 2 the polynomial? 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two SOLUTION: If a polynomial divided by a binomial has no polynomials is . remainder, then the polynomial has two factors: the binomial and the quotient. ANSWER: ANSWER: 5-2 Dividing Polynomials Sample answer: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems 47. CCSS ARGUMENTS Identify the expression that in this lesson. What is the relationship between the does not belong with the other three. Explain your degrees of the dividend, the divisor, and the quotient? reasoning. SOLUTION: Sample answer: For example:

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the denominator of contains a variable, it is not a polynomial.

ANSWER:

In this exercise, the degree of the dividend is 2. The does not belong with the other three. The other degree of the divisor and the quotient are each 1. three expressions are polynomials. Since the The degree of the quotient plus the degree of the divisor equals the degree of the dividend. denominator of contains a variable, it is not a ANSWER: polynomial. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the 48. WRITING IN MATH Use the information at the dividend. beginning of the lesson to write assembly instruction using the division of polynomials to make a paper 46. OPEN ENDED Write a quotient of two polynomials cover for your textbook.

for which the remainder is 3. SOLUTION: SOLUTION: Sample answer: By dividing the area of the paper Sample answer: Begin by multiplying two binomials 140x2 + 60x by the height of the book jacket 10x, the such as (x + 2)(x + 3) which simplifies to x2 + 5x + quotient of 14x + 6 provides the length of the book 6. In order to get a remainder of 3 when divided, add jacket. The front and back cover are each 6x units 2 long and the spine is 2x units long. Then, subtracting 3 to the trinomial to get x + 5x + 9. When divided, 14x, we are left with 6 inches. Half of this length is there will be a remainder of 3. The quotient of two the width of each flap.

polynomials is . ANSWER: 2 ANSWER: Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book

Sample answer: jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

47. CCSS ARGUMENTS Identify the expression that 49. An office employs x women and 3 men. What is the does not belong with the other three. Explain your ratio of the total number of employees to the number reasoning. of women?

A eSolutions Manual - Powered by Cognero B Page 12

C SOLUTION:

does not belong with the other three. The other D

three expressions are polynomials. Since the SOLUTION: denominator of contains a variable, it is not a Total number of employees: x + 3 Number of women: x polynomial. Ratio of the total number of employees to the number of women is: ANSWER: does not belong with the other three. The other The correct choice is A. three expressions are polynomials. Since the ANSWER: denominator of contains a variable, it is not a A polynomial. 50. SAT/ACT Which polynomial has degree 3? 3 2 4 48. WRITING IN MATH Use the information at the A x + x –2x 2 beginning of the lesson to write assembly instruction B –2x – 3x + 4 using the division of polynomials to make a paper C 3x – 3 cover for your textbook. D x2 + x + 123 3 SOLUTION: E 1 + x + x Sample answer: By dividing the area of the paper SOLUTION: 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book To determine the degree of a polynomial, look at the jacket. The front and back cover are each 6x units term with the greatest exponent. long and the spine is 2x units long. Then, subtracting The correct choice is E. 14x, we are left with 6 inches. Half of this length is the width of each flap. ANSWER: E ANSWER: 2 51. GRIDDED RESPONSE In the figure below, Sample answer: By dividing 140x + 60x by 10x, the

quotient of 14x + 6 provides the length of the book m + n + p = ? jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women? SOLUTION: A Sum of the exterior angles of a triangle is . So, B ANSWER: C 360

D 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x SOLUTION: 2 H –10x + 17x Total number of employees: x + 3 2 Number of women: x G –10x 2 Ratio of the total number of employees to the number J 2x + 17x of women is: SOLUTION:

The correct choice is A.

ANSWER: A The correct choice is H.

50. SAT/ACT Which polynomial has degree 3? ANSWER: A x3 + x2 –2x4 H 2 B –2x – 3x + 4 Simplify. 3 2 3 C 3x – 3 53. (5x + 2x – 3x + 4) – (2x – 4x) D x2 + x + 123 3 SOLUTION: E 1 + x + x Use the Distributive Property and then combine like SOLUTION: terms. To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is E.

ANSWER: E ANSWER: 51. GRIDDED RESPONSE In the figure below, 3 2 m + n + p = ? 3x + 2x + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

SOLUTION: Sum of the exterior angles of a triangle is . ANSWER: So, 2y 3 + 3y 2 – 9y + 8

ANSWER: 55. 4a(2a – 3) + 3a(5a – 4) 360 SOLUTION: 2 2 Use the Distributive Property and then combine like 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = terms. 2 F 2x H –10x2 + 17x G 2 –10x ANSWER: 2 J 2x + 17x 2 23a – 24a SOLUTION: 56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms.

The correct choice is H.

ANSWER: ANSWER: H 3 2 2 3 2c – 3c d – 2cd + 3d

Simplify. 2 2 3 3 2 3 57. (xy) (2xy z) 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: SOLUTION: Apply the properties of exponents to simplify the Use the Distributive Property and then combine like expression. terms.

ANSWER: 8x5y 8z 3

2 –2 2 2 ANSWER: 58. (3ab ) (2a b) 3 2 3x + 2x + x + 4 SOLUTION:

3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION:

ANSWER: ANSWER:

2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) 59. LANDSCAPING Amado wants to plant a garden SOLUTION: and surround it with decorative stones. He has Use the Distributive Property and then combine like enough stones to enclose a rectangular garden with a terms. perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be?

ANSWER: SOLUTION: 2 Let x be the length and y be the width of the garden. 23a – 24a 56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms.

ANSWER: The solution of the inequality is 2c3 – 3c2d – 2cd 2 + 3d 3 . Since y represent the 2 2 3 57. (xy) (2xy z) width of the garden and the sum of the length and width is 34, the width of the garden is SOLUTION: Apply the properties of exponents to simplify the expression. ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 2 ANSWER: 60. x + 6x + 2 = 0 8x5y 8z 3 SOLUTION:

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER:

–3 ±

59. LANDSCAPING Amado wants to plant a garden 2 and surround it with decorative stones. He has 61. x – 8x – 3 = 0 enough stones to enclose a rectangular garden with a SOLUTION: perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

62. 2x2 + 6x + 5 = 0 SOLUTION: The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0

SOLUTION: ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

63.

SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the ANSWER: quadratic function lies between –6 and –5, and between –3 and –2. –3 ± ANSWER: 2 61. x – 8x – 3 = 0 between –6 and –5; between –3 and –2 SOLUTION:

64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3. ANSWER: ANSWER:

between 0 and 1; between 2 and 3

62. 2x2 + 6x + 5 = 0

SOLUTION: 65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days ANSWER: per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will State the consecutive integers between which maximize income. Then find the maximum income. the zeros of each quadratic function are located. SOLUTION: Let M be the lawns mowed and L be the small 63. landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works SOLUTION: 10 hours per day, 5 days a week, he can do at most The sign of f (x) changes between the x values –6 20 mowing jobs per day or 100 per week. So, M ≤ and –5, and between –3 and –2. So the zeros of the 100. quadratic function lies between –6 and –5, and between –3 and –2. If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total ANSWER: of 10 hours per day. So, for the week, if he does 15 between –6 and –5; between –3 and –2 landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that relates the number and charge for each type of job. 64. Since he charges $35 for each lawn mowed and SOLUTION: $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and The sign of f (x) changes between the x values L to determine the earnings for each combination of between 0 and 1, and between 2 and 3. So the zeros jobs. of the quadratic function lies between 0 and 1, and between 2 and 3. (M, L) F(M, L) = 35M + F(M, L) ANSWER: 125L between 0 and 1; between 2 and 3 (100, 0) F(M, L) = 35(100) + 3500 125(0)

65. (55, 15) F(M, L) = 35(55) + 125 3800 SOLUTION: (15) The sign of f (x) changes between the x values (0, 15) F(M, L) = 35(0) + 125 1875 between –1 and 0, and between 2 and 3. So the (15) zeros of the quadratic function lies between –1 and 0, and between 2 and 3. 15 landscape jobs and 55 lawns; $3800 ANSWER: between –1 and 0; between 2 and 3 ANSWER: 15 landscape jobs and 55 lawns; $3800 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 Find each value if f (x) = 4x + 3, g(x) = –x2, and minutes. He works at most 10 hours per day, 5 days h(x) = –2x2 – 2x + 4. per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape 67. f (–6) jobs per day. Find the combination of lawns mowed SOLUTION: and completed landscape jobs per week that will Substitute –6 for x in f (x). maximize income. Then find the maximum income. SOLUTION: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most ANSWER: 20 mowing jobs per day or 100 per week. So, M ≤ –21 100.

If the landscaper does 3 landscaping jobs per day, he 68. g(–8) can only do 11 mowing jobs per day and work a total SOLUTION: of 10 hours per day. So, for the week, if he does 15 Substitute –8 for x in g(x). landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M ANSWER: + 125L. Next, substitute in the key values of M and –64 L to determine the earnings for each combination of jobs. 69. h(3) SOLUTION: (M, L) F(M, L) = 35M + F(M, L) Substitute 3 for x in h(x). 125L

(100, 0) F(M, L) = 35(100) + 3500 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 (15) ANSWER: (0, 15) F(M, L) = 35(0) + 125 1875 –20 (15) 70. f (c)

15 landscape jobs and 55 lawns; $3800 SOLUTION: Substitute c for x in f (x). ANSWER: 15 landscape jobs and 55 lawns; $3800 Find each value if f (x) = 4x + 3, g(x) = –x2, and h(x) = –2x2 – 2x + 4. ANSWER: 67. f (–6) 4c + 3 SOLUTION: 71. g(3d) Substitute –6 for x in f (x). SOLUTION: Substitute 3d for x in g(x).

ANSWER: –21 ANSWER: 2 –9d 68. g(–8) 72. h(2b + 1) SOLUTION: Substitute –8 for x in g(x). SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –64 69. h(3) ANSWER: SOLUTION: –8b2 – 12b Substitute 3 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) Simplify. SOLUTION: 1.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) ANSWER:

SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION: ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

ANSWER: 2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: 5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

D. ANSWER: SOLUTION: First rewrite the divisor so the x-term is first. Then use long division. 5 2 6. (y – 3y – 20) ÷ (y – 2)

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

D. ANSWER:

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division. 2 9. (18a + 6a + 9) ÷ (3a – 2)

SOLUTION:

The correct choice is A. ANSWER:

ANSWER: A

Simplify. 10. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: SOLUTION:

ANSWER:

11.

2 SOLUTION: 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: ANSWER: 3y + 5 Simplify

12.

10. SOLUTION:

SOLUTION:

ANSWER: 3a2b – 2ab2

13.

SOLUTION:

ANSWER:

11. ANSWER: SOLUTION: x + 3y –2

14.

SOLUTION:

ANSWER: ANSWER: 3y + 5 Simplify

12. 15.

SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

13. ANSWER: 2 2a + b – 3 SOLUTION:

16.

SOLUTION:

ANSWER: x + 3y –2

14. ANSWER: 4c2d2 – 6 SOLUTION:

17.

SOLUTION:

ANSWER:

15. ANSWER: 3np – 6 + 7p SOLUTION: 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb ANSWER: is: 2 2a + b – 3

16. ANSWER: SOLUTION: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: ANSWER: The average number of cookies produced per worker 2 2 4c d – 6 is:

17.

SOLUTION: ANSWER:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) ANSWER: SOLUTION: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. ANSWER: SOLUTION: The average amount of energy saved per CFL bulb is:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced ANSWER: per worker. SOLUTION: The average number of cookies produced per worker 4 3 2 –1 is: 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

ANSWER: z3 – 2z2 – 4 Simplify. 2 5 3 2 20. (a – 8a – 26) ÷ (a + 2) 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER: ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) 24. SOLUTION: SOLUTION:

ANSWER: ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) 4 2 SOLUTION: 25. (g – 3g – 18) ÷ (g – 2) SOLUTION:

ANSWER: z3 – 2z2 – 4 ANSWER: 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: 2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER:

27.

4 2 25. (g – 3g – 18) ÷ (g – 2) SOLUTION: SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27.

SOLUTION: ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

28. 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

Synthetic division:

ANSWER:

ANSWER: 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: 32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 Synthetic division: be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER:

The depressed polynomial is . 6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION:

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. ANSWER: SOLUTION:

34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. The depressed polynomial is . • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b Since the volume of the box is the product of length, . Explain the process you used to discover the

width and height, the width and length of box are (2x answer. + 3)(3x + 5). SOLUTION: a. Let x be the number. ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a b. Sample answer: Let x be the number. Multiply the generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If x by 4 to get 4x. Then add x +15 to the product to get the current of the generator is I = t + 4, write an 5x + 15. Divide the polynomial by x + 3. The quotient expression that represents the voltage. is 5. SOLUTION: ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + Synthetic division: 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is

the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

ANSWER: a. Perform the division indicated by . 2 V(t) = t + 5t + 6 b. About how many subscriptions will be sold if 34. ENTERTAINMENT A magician gives these $1500 is spent on advertising? instructions to a volunteer. SOLUTION: • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. a. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION:

a. Let x be the number. b. There are 15 hundreds in 1500. Substitute a = 15.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: a.5 Therefore, about 2423 subscriptions will be sold. b. Sample answer: Let x be the number. ANSWER: Multiply the x by 4 to get 4x. Then add x + 15 to the

product to get 5x + 15. Divide the polynomial by x + a. 3. The quotient is 5. b. about 2423 subscriptions 35. BUSINESS The number of magazine subscriptions Simplify. sold can be estimated by ,where a is 36. (x4 – y4) ÷ (x – y) the amount of money the company spent on SOLUTION: advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: ANSWER: 2 2 (x + y )(x + y) a. 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

Therefore, about 2423 subscriptions will be sold. ANSWER:

ANSWER:

b. about 2423 subscriptions 39.

Simplify. SOLUTION: 36. (x4 – y4) ÷ (x – y) SOLUTION:

ANSWER: n2 – n – 1 ANSWER: 2 2 (x + y )(x + y) 40.

3 2 2 37. (28c d – 21cd ) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION: 41.

SOLUTION:

ANSWER:

39. ANSWER: SOLUTION: 3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent ANSWER: the model. n2 – n – 1 c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? 40. SOLUTION: a. SOLUTION:

ANSWER:

The width is x + 3. 41. b. SOLUTION:

ANSWER: Yes, the concrete model checks with the algebraic model. 3z 4 – z 3 + 2 z2 – 4z + 9– ANSWER: a. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION:

a. The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

The width is x + 3. 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the b. remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. c. SOLUTION: Use synthetic division.

Yes, the concrete model checks with the algebraic model. The remainder is 26. So, Sharon is correct. Jamal ANSWER: actually divided by x + 3. a. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

The width is x + 3. binomial and the quotient. ANSWER: 2 b. 2x + 7x + 3 ÷ (2x + 1) The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the c. degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example: yes

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The remainder is 26. So, Sharon is correct. Jamal The degree of the quotient plus the degree of the actually divided by x + 3. divisor equals the degree of the dividend.

ANSWER: ANSWER: Sample answer: Sharon; Jamal actually divided by x Sample answer: The degree of the quotient plus the + 3. degree of the divisor equals the degree of the dividend. 44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell 46. OPEN ENDED Write a quotient of two polynomials you about the relationship between the binomial and for which the remainder is 3. the polynomial? SOLUTION: SOLUTION: Sample answer: Begin by multiplying two binomials If a polynomial divided by a binomial has no such as (x + 2)(x + 3) which simplifies to x2 + 5x + remainder, then the polynomial has two factors: the 6. In order to get a remainder of 3 when divided, add binomial and the quotient. 2 3 to the trinomial to get x + 5x + 9. When divided, ANSWER: there will be a remainder of 3. The quotient of two The binomial is a factor of the polynomial. polynomials is . 45. REASONING Review any of the division problems in this lesson. What is the relationship between the ANSWER: degrees of the dividend, the divisor, and the quotient? Sample answer: SOLUTION: Sample answer: For example: 47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the

In this exercise, the degree of the dividend is 2. The denominator of contains a variable, it is not a degree of the divisor and the quotient are each 1.

The degree of the quotient plus the degree of the polynomial. divisor equals the degree of the dividend. ANSWER:

ANSWER: does not belong with the other three. The other Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the three expressions are polynomials. Since the dividend. denominator of contains a variable, it is not a 46. OPEN ENDED Write a quotient of two polynomials polynomial. for which the remainder is 3. SOLUTION: 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction Sample answer: Begin by multiplying two binomials 2 using the division of polynomials to make a paper such as (x + 2)(x + 3) which simplifies to x + 5x + cover for your textbook. 6. In order to get a remainder of 3 when divided, add 2 SOLUTION: 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two Sample answer: By dividing the area of the paper 2 140x + 60x by the height of the book jacket 10x, the polynomials is . quotient of 14x + 6 provides the length of the book jacket. The front and back cover are each 6x units ANSWER: long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is Sample answer: the width of each flap. ANSWER: 47. CCSS ARGUMENTS Identify the expression that 2 Sample answer: By dividing 140x + 60x by 10x, the does not belong with the other three. Explain your quotient of 14x + 6 provides the length of the book reasoning. jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women?

SOLUTION: A

does not belong with the other three. The other B three expressions are polynomials. Since the C denominator of contains a variable, it is not a

polynomial. D

ANSWER: SOLUTION: does not belong with the other three. The other Total number of employees: x + 3 Number of women: x three expressions are polynomials. Since the Ratio of the total number of employees to the number of women is: denominator of contains a variable, it is not a

polynomial. The correct choice is A. 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction ANSWER: using the division of polynomials to make a paper A cover for your textbook. 50. SAT/ACT Which polynomial has degree 3? SOLUTION: A x3 + x2 –2x4 Sample answer: By dividing the area of the paper 2 2 B 140x + 60x by the height of the book jacket 10x, the –2x – 3x + 4 quotient of 14x + 6 provides the length of the book C 3x – 3 2 3 jacket. The front and back cover are each 6x units D x + x + 12 3 long and the spine is 2x units long. Then, subtracting E 1 + x + x 14x, we are left with 6 inches. Half of this length is the width of each flap. SOLUTION: To determine the degree of a polynomial, look at the ANSWER: term with the greatest exponent. 2 Sample answer: By dividing 140x + 60x by 10x, the The correct choice is E. quotient of 14x + 6 provides the length of the book 5-2 Dividingjacket. Then, Polynomials subtracting 14x, we are left with 6 ANSWER: inches. Half of this length is the width of each flap. E

49. An office employs x women and 3 men. What is the 51. GRIDDED RESPONSE In the figure below, ratio of the total number of employees to the number m + n + p = ? of women?

C SOLUTION: Sum of the exterior angles of a triangle is . D So,

SOLUTION: ANSWER: Total number of employees: x + 3 360 Number of women: x Ratio of the total number of employees to the number 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = of women is: 2 F 2x

H –10x2 + 17x 2 The correct choice is A. G –10x J 2x2 + 17x ANSWER: A SOLUTION:

50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 2 B –2x – 3x + 4 C 3x – 3 The correct choice is H. 2 3 D x + x + 12 ANSWER: 3 E 1 + x + x H SOLUTION: Simplify. 3 2 3 To determine the degree of a polynomial, look at the 53. (5x + 2x – 3x + 4) – (2x – 4x) term with the greatest exponent. The correct choice is E. SOLUTION: Use the Distributive Property and then combine like ANSWER: terms. E

51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 3x3 + 2x2 + x + 4

SOLUTION: 3 2 54. (2y – 3y + 8) + (3y – 6y) Sum of the exterior angles of a triangle is . So, SOLUTION: eSolutions Manual - Powered by Cognero Page 13 ANSWER: 360 ANSWER: 2 2 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2y 3 + 3y 2 – 9y + 8 2 F 2x 2 55. 4a(2a – 3) + 3a(5a – 4) H –10x + 17x 2 G –10x SOLUTION: J 2x2 + 17x Use the Distributive Property and then combine like terms. SOLUTION:

ANSWER: 2 23a – 24a The correct choice is H. 56. (c + d)(c – d)(2c – 3d) ANSWER: SOLUTION: H Use the FOIL method and then combine like terms. Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: ANSWER: Use the Distributive Property and then combine like 2c3 – 3c2d – 2cd 2 + 3d 3 terms. 2 2 3 57. (xy) (2xy z)

SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 3 2 3x + 2x + x + 4 ANSWER: 8x5y 8z 3 3 2 54. (2y – 3y + 8) + (3y – 6y) 2 –2 2 2 SOLUTION: 58. (3ab ) (2a b) SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: ANSWER: Use the Distributive Property and then combine like terms.

59. LANDSCAPING Amado wants to plant a garden ANSWER: and surround it with decorative stones. He has 2 enough stones to enclose a rectangular garden with a 23a – 24a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the 56. (c + d)(c – d)(2c – 3d) width of his garden be? SOLUTION: SOLUTION: Use the FOIL method and then combine like terms. Let x be the length and y be the width of the garden.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION: Apply the properties of exponents to simplify the The solution of the inequality is expression. . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: ANSWER: 8x5y 8z 3 0 to 10 ft or 24 to 34 ft 2 –2 2 2 58. (3ab ) (2a b) Solve each equation by completing the square. 2 SOLUTION: 60. x + 6x + 2 = 0 SOLUTION:

ANSWER:

59. LANDSCAPING Amado wants to plant a garden

and surround it with decorative stones. He has

enough stones to enclose a rectangular garden with a ANSWER:

perimeter of 68 feet, but he wants the garden to –3 ± cover no more than 240 square feet. What could the width of his garden be? 2 61. x – 8x – 3 = 0 SOLUTION: Let x be the length and y be the width of the garden. SOLUTION:

ANSWER: The solution of the inequality is . Since y represent the width of the garden and the sum of the length and 62. 2x2 + 6x + 5 = 0 width is 34, the width of the garden is SOLUTION:

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

ANSWER: 63. –3 ± SOLUTION: The sign of f (x) changes between the x values –6 2 61. x – 8x – 3 = 0 and –5, and between –3 and –2. So the zeros of the SOLUTION: quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: between –6 and –5; between –3 and –2

64. ANSWER: SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros 62. 2x2 + 6x + 5 = 0 of the quadratic function lies between 0 and 1, and between 2 and 3. SOLUTION: ANSWER: between 0 and 1; between 2 and 3

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: ANSWER: between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 State the consecutive integers between which minutes. He works at most 10 hours per day, 5 days the zeros of each quadratic function are located. per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape 63. jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will SOLUTION: maximize income. Then find the maximum income. The sign of f (x) changes between the x values –6 SOLUTION: and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and Let M be the lawns mowed and L be the small between –3 and –2. landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works ANSWER: 10 hours per day, 5 days a week, he can do at most between –6 and –5; between –3 and –2 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he 64. can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 SOLUTION: landscaping jobs, he can mow 55 lawns. The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros To maximize the earnings, write a function that of the quadratic function lies between 0 and 1, and relates the number and charge for each type of job. between 2 and 3. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M ANSWER: + 125L. Next, substitute in the key values of M and between 0 and 1; between 2 and 3 L to determine the earnings for each combination of jobs.

65. (M, L) F(M, L) = 35M + F(M, L) 125L SOLUTION: The sign of f (x) changes between the x values (100, 0) F(M, L) = 35(100) + 3500 between –1 and 0, and between 2 and 3. So the 125(0) zeros of the quadratic function lies between –1 and 0, and between 2 and 3. (55, 15) F(M, L) = 35(55) + 125 3800 (15) ANSWER: between –1 and 0; between 2 and 3 (0, 15) F(M, L) = 35(0) + 125 1875 (15) 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days 15 landscape jobs and 55 lawns; $3800 per week. He earns $35 per lawn and $125 per ANSWER: landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed 15 landscape jobs and 55 lawns; $3800 and completed landscape jobs per week that will 2 maximize income. Then find the maximum income. Find each value if f (x) = 4x + 3, g(x) = –x , and h(x) = –2x2 – 2x + 4. SOLUTION: 67. f (–6) Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most SOLUTION: 3 landscaping jobs per day, L ≤ 15. Since he works Substitute –6 for x in f (x). 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total ANSWER: of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. –21

To maximize the earnings, write a function that 68. g(–8) relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and SOLUTION: $125 for each small landscaping job, F(M, L) = 35M Substitute –8 for x in g(x). + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) ANSWER: 125L –64

(100, 0) F(M, L) = 35(100) + 3500 69. h(3) 125(0) SOLUTION: Substitute 3 for x in h(x). (55, 15) F(M, L) = 35(55) + 125 3800 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800 ANSWER: ANSWER: –20 15 landscape jobs and 55 lawns; $3800 70. f (c) 2 Find each value if f (x) = 4x + 3, g(x) = –x , and SOLUTION: 2 h(x) = –2x – 2x + 4. Substitute c for x in f (x). 67. f (–6) SOLUTION: Substitute –6 for x in f (x).

ANSWER:

4c + 3 71. g(3d) ANSWER: SOLUTION: –21 Substitute 3d for x in g(x).

68. g(–8) SOLUTION: Substitute –8 for x in g(x). ANSWER: 2 –9d

72. h(2b + 1)

ANSWER: SOLUTION: –64 Substitute 2b + 1 for x in h(x).

69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –8b2 – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1)

Simplify. SOLUTION:

SOLUTION:

ANSWER: 4y + 2x – 2 ANSWER: 2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION: 5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER: ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) 5 2 SOLUTION: 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

ANSWER: SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

C. ANSWER:

SOLUTION: 2 First rewrite the divisor so the x-term is first. Then 9. (18a + 6a + 9) ÷ (3a – 2) use long division. SOLUTION:

ANSWER: The correct choice is A.

ANSWER: A 10. Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: SOLUTION:

ANSWER:

ANSWER: 11.

SOLUTION: 2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER: 3y + 5 ANSWER: Simplify

12.

SOLUTION: 10.

SOLUTION: ANSWER: 3a2b – 2ab2

13.

SOLUTION:

ANSWER:

11. ANSWER: x + 3y –2 SOLUTION:

14.

SOLUTION:

ANSWER: ANSWER: 3y + 5

Simplify 15.

12. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

ANSWER: 13. 2 2a + b – 3

SOLUTION: 16.

SOLUTION:

ANSWER: x + 3y –2

ANSWER: 14. 4c2d2 – 6

SOLUTION: 17.

SOLUTION:

ANSWER:

15. 3np – 6 + 7p ENERGY SOLUTION: 18. Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is: ANSWER: 2 2a + b – 3

16. ANSWER: –b + 8 SOLUTION: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker ANSWER: is: 4c2d2 – 6

17.

SOLUTION: ANSWER:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: ANSWER: 3np – 6 + 7p

SOLUTION: The average amount of energy saved per CFL bulb

is: 3 2 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + ANSWER: 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker.

SOLUTION: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) The average number of cookies produced per worker is: SOLUTION:

ANSWER: ANSWER: z3 – 2z2 – 4

Simplify. 5 3 2 2 23. (x – 4x + 4x ) ÷ (x – 4) 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: SOLUTION:

ANSWER: ANSWER:

24. 21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: SOLUTION:

ANSWER: ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) 25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION: SOLUTION:

ANSWER: ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4)

2 SOLUTION: 26. (6a – 3a + 9) ÷ (3a – 2)

SOLUTION:

ANSWER:

24.

SOLUTION:

ANSWER:

27.

SOLUTION: 4 2 25. (g – 3g – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

27. ANSWER: SOLUTION:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER: ANSWER:

28. 6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: SOLUTION:

Synthetic division:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION: 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION:

Synthetic division:

ANSWER: ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) 32. CCSS REASONING A rectangular box for a new SOLUTION: product is designed in such a way that the three dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is Synthetic division: always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division.

ANSWER:

The depressed polynomial is .

6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION:

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION: ANSWER:

32. CCSS REASONING A rectangular box for a new

product is designed in such a way that the three Synthetic division: dimensions always have a particular relationship defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division. ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the The depressed polynomial is . product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the

answer. Since the volume of the box is the product of length, width and height, the width and length of box are (2x SOLUTION: + 3)(3x + 5). a. Let x be the number.

ANSWER: 2x + 3, 3x + 5

33. PHYSICS The voltage V is related to current I and b. Sample answer: Let x be the number. Multiply the power P by the equation . The power of a x by 4 to get 4x. Then add x +15 to the product to get 3 2 5x + 15. Divide the polynomial by x + 3. The quotient generator is modeled by P(t) = t + 9t + 26t + 24. If is 5. the current of the generator is I = t + 4, write an expression that represents the voltage. ANSWER: SOLUTION: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. Synthetic division: 35. BUSINESS The number of magazine subscriptions

sold can be estimated by ,where a is the amount of money the company spent on advertising in hundreds of dollars and n is the number

of subscriptions sold.

a. Perform the division indicated by . ANSWER: 2 b. About how many subscriptions will be sold if V(t) = t + 5t + 6 $1500 is spent on advertising? 34. ENTERTAINMENT A magician gives these SOLUTION: instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the a. product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION: b. There are 15 hundreds in 1500. Substitute a = 15. a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: Therefore, about 2423 subscriptions will be sold. a. 5 ANSWER: b. Sample answer: Let x be the number.

Multiply the x by 4 to get 4x. Then add x + 15 to the a. product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. b. about 2423 subscriptions

35. BUSINESS The number of magazine subscriptions Simplify. 36. (x4 – y4) ÷ (x – y) sold can be estimated by ,where a is SOLUTION: the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? ANSWER: 2 2 SOLUTION: (x + y )(x + y)

a. 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

ANSWER: b. There are 15 hundreds in 1500. Substitute a = 15.

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER: Therefore, about 2423 subscriptions will be sold. ANSWER:

a. 39. b. about 2423 subscriptions SOLUTION: Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION:

ANSWER: n2 – n – 1

ANSWER: 2 2 40. (x + y )(x + y)

SOLUTION: 3 2 2 37. (28c d – 21cd ) ÷ (14cd) SOLUTION:

ANSWER: ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION: 41. SOLUTION:

ANSWER:

39. ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9– SOLUTION:

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. ANSWER: c. NUMERICAL Solve this problem algebraically n2 – n – 1 using synthetic or long division. Does your concrete model check with your algebraic model?

40. SOLUTION: a. SOLUTION:

ANSWER:

The width is x + 3.

41. b.

SOLUTION: c.

Yes, the concrete model checks with the algebraic ANSWER: model.

3z 4 – z 3 + 2 z2 – 4z + 9– ANSWER: a. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model?

SOLUTION: a. The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 The width is x + 3. 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – b. 100. Is either of them correct? Explain your reasoning. SOLUTION: c. Use synthetic division.

Yes, the concrete model checks with the algebraic model. The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3. ANSWER: a. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

The width is x + 3. ANSWER: The binomial is a factor of the polynomial. b. 2x 2 + 7x + 3 ÷ (2x + 1) 45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? c. SOLUTION: Sample answer: For example:

yes 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. SOLUTION: Use synthetic division.

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the The remainder is 26. So, Sharon is correct. Jamal divisor equals the degree of the dividend. actually divided by x + 3. ANSWER: ANSWER: Sample answer: The degree of the quotient plus the Sample answer: Sharon; Jamal actually divided by x degree of the divisor equals the degree of the + 3. dividend.

44. CHALLENGE If a polynomial is divided by a 46. OPEN ENDED Write a quotient of two polynomials binomial and the remainder is 0, what does this tell for which the remainder is 3. you about the relationship between the binomial and SOLUTION: the polynomial? Sample answer: Begin by multiplying two binomials SOLUTION: such as (x + 2)(x + 3) which simplifies to x2 + 5x + If a polynomial divided by a binomial has no 6. In order to get a remainder of 3 when divided, add remainder, then the polynomial has two factors: the 2 3 to the trinomial to get x + 5x + 9. When divided, binomial and the quotient. there will be a remainder of 3. The quotient of two ANSWER: polynomials is . The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems ANSWER: in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? Sample answer: SOLUTION: 47. CCSS ARGUMENTS Identify the expression that Sample answer: For example: does not belong with the other three. Explain your

reasoning.

SOLUTION:

does not belong with the other three. The other three expressions are polynomials. Since the

denominator of contains a variable, it is not a In this exercise, the degree of the dividend is 2. The polynomial. degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the ANSWER: divisor equals the degree of the dividend. does not belong with the other three. The other ANSWER: Sample answer: The degree of the quotient plus the three expressions are polynomials. Since the degree of the divisor equals the degree of the denominator of contains a variable, it is not a dividend. polynomial. 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction SOLUTION: using the division of polynomials to make a paper Sample answer: Begin by multiplying two binomials cover for your textbook. such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add SOLUTION: 2 Sample answer: By dividing the area of the paper 3 to the trinomial to get x + 5x + 9. When divided, 2 there will be a remainder of 3. The quotient of two 140x + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book polynomials is . jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting ANSWER: 14x, we are left with 6 inches. Half of this length is the width of each flap.

Sample answer: ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the 47. CCSS ARGUMENTS Identify the expression that quotient of 14x + 6 provides the length of the book does not belong with the other three. Explain your jacket. Then, subtracting 14x, we are left with 6 reasoning. inches. Half of this length is the width of each flap.

49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number of women?

A SOLUTION:

B does not belong with the other three. The other

three expressions are polynomials. Since the C denominator of contains a variable, it is not a D polynomial. SOLUTION: ANSWER: Total number of employees: x + 3 does not belong with the other three. The other Number of women: x Ratio of the total number of employees to the number three expressions are polynomials. Since the of women is:

denominator of contains a variable, it is not a polynomial. The correct choice is A.

48. WRITING IN MATH Use the information at the ANSWER: beginning of the lesson to write assembly instruction A using the division of polynomials to make a paper cover for your textbook. 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 SOLUTION: 2 Sample answer: By dividing the area of the paper B –2x – 3x + 4 C 3x – 3 140x2 + 60x by the height of the book jacket 10x, the 2 3 quotient of 14x + 6 provides the length of the book D x + x + 12 3 jacket. The front and back cover are each 6x units E 1 + x + x long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is SOLUTION: the width of each flap. To determine the degree of a polynomial, look at the term with the greatest exponent. ANSWER: The correct choice is E. 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book ANSWER: jacket. Then, subtracting 14x, we are left with 6 E inches. Half of this length is the width of each flap. 51. GRIDDED RESPONSE In the figure below, 49. An office employs x women and 3 men. What is the m + n + p = ? ratio of the total number of employees to the number of women?

B SOLUTION: C Sum of the exterior angles of a triangle is . So, D ANSWER: SOLUTION: 360 Total number of employees: x + 3 2 2 Number of women: x 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2 Ratio of the total number of employees to the number F 2x of women is: H –10x2 + 17x 2 G –10x 2 The correct choice is A. J 2x + 17x

ANSWER: SOLUTION: A

50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 2 B –2x – 3x + 4 The correct choice is H. C 3x – 3 ANSWER: 2 3 D x + x + 12 H 3 E 1 + x + x Simplify. 3 2 3 SOLUTION: 53. (5x + 2x – 3x + 4) – (2x – 4x) To determine the degree of a polynomial, look at the SOLUTION: term with the greatest exponent. The correct choice is E. Use the Distributive Property and then combine like terms. ANSWER: E

51. GRIDDED RESPONSE In the figure below, m + n + p = ?

ANSWER: 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) SOLUTION: Sum of the exterior angles of a triangle is . SOLUTION: So,

ANSWER: 360 ANSWER: 2y 3 + 3y 2 – 9y + 8 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x 55. 4a(2a – 3) + 3a(5a – 4) H –10x2 + 17x SOLUTION: 2 G –10x Use the Distributive Property and then combine like terms. J 2 2x + 17x SOLUTION:

ANSWER: 2 23a – 24a

The correct choice is H. 56. (c + d)(c – d)(2c – 3d) SOLUTION: ANSWER: Use the FOIL method and then combine like terms. H Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) ANSWER: SOLUTION: 2c3 – 3c2d – 2cd 2 + 3d 3 Use the Distributive Property and then combine like 2 2 3 terms. 57. (xy) (2xy z)

SOLUTION:

Apply the properties of exponents to simplify the expression.

ANSWER: ANSWER: 5-2 Dividing3x3 + 2x2 Polynomials + x + 4 8x5y 8z 3

3 2 2 –2 2 2 54. (2y – 3y + 8) + (3y – 6y) 58. (3ab ) (2a b) SOLUTION: SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) ANSWER: SOLUTION: Use the Distributive Property and then combine like terms. 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a ANSWER: perimeter of 68 feet, but he wants the garden to 2 23a – 24a cover no more than 240 square feet. What could the width of his garden be? 56. (c + d)(c – d)(2c – 3d) SOLUTION: SOLUTION: Let x be the length and y be the width of the garden. Use the FOIL method and then combine like terms.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

2 2 3 57. (xy) (2xy z) SOLUTION: The solution of the inequality is Apply the properties of exponents to simplify the . Since y represent the expression. width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: ANSWER: 0 to 10 ft or 24 to 34 ft 8x5y 8z 3 Solve each equation by completing the square. 2 –2 2 2 58. (3ab ) (2a b) 60. x2 + 6x + 2 = 0 SOLUTION: SOLUTION:

ANSWER: eSolutions Manual - Powered by Cognero Page 14

59. LANDSCAPING Amado wants to plant a garden ANSWER: and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a –3 ± perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the 2 width of his garden be? 61. x – 8x – 3 = 0 SOLUTION: SOLUTION: Let x be the length and y be the width of the garden.

ANSWER:

The solution of the inequality is . Since y represent the 62. 2x2 + 6x + 5 = 0 width of the garden and the sum of the length and width is 34, the width of the garden is SOLUTION:

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION:

ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

63. ANSWER:

–3 ± SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the 2 61. x – 8x – 3 = 0 quadratic function lies between –6 and –5, and between –3 and –2. SOLUTION: ANSWER: between –6 and –5; between –3 and –2

64. SOLUTION: ANSWER: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3. 62. 2x2 + 6x + 5 = 0 SOLUTION: ANSWER: between 0 and 1; between 2 and 3

65. SOLUTION: The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3 ANSWER: 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days State the consecutive integers between which per week. He earns $35 per lawn and $125 per the zeros of each quadratic function are located. landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will 63. maximize income. Then find the maximum income. SOLUTION: SOLUTION: The sign of f (x) changes between the x values –6 Let M be the lawns mowed and L be the small and –5, and between –3 and –2. So the zeros of the landscaping jobs. Since the landscaper will do at most quadratic function lies between –6 and –5, and 3 landscaping jobs per day, L 15. Since he works between –3 and –2. ≤ 10 hours per day, 5 days a week, he can do at most ANSWER: 20 mowing jobs per day or 100 per week. So, M ≤ 100. between –6 and –5; between –3 and –2

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 64. landscaping jobs, he can mow 55 lawns. SOLUTION: The sign of f (x) changes between the x values To maximize the earnings, write a function that between 0 and 1, and between 2 and 3. So the zeros relates the number and charge for each type of job. of the quadratic function lies between 0 and 1, and Since he charges $35 for each lawn mowed and between 2 and 3. $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and ANSWER: L to determine the earnings for each combination of between 0 and 1; between 2 and 3 jobs.

(M, L) F(M, L) = 35M + F(M, L) 125L 65. SOLUTION: (100, 0) F(M, L) = 35(100) + 3500 The sign of f (x) changes between the x values 125(0) between –1 and 0, and between 2 and 3. So the zeros of the quadratic function lies between –1 and 0, (55, 15) F(M, L) = 35(55) + 125 3800 and between 2 and 3. (15)

ANSWER: (0, 15) F(M, L) = 35(0) + 125 1875 between –1 and 0; between 2 and 3 (15)

66. BUSINESS A landscaper can mow a lawn in 30 15 landscape jobs and 55 lawns; $3800 minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days ANSWER: per week. He earns $35 per lawn and $125 per 15 landscape jobs and 55 lawns; $3800 landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed Find each value if f (x) = 4x + 3, g(x) = –x2, and and completed landscape jobs per week that will 2 maximize income. Then find the maximum income. h(x) = –2x – 2x + 4. 67. f (–6) SOLUTION: Let M be the lawns mowed and L be the small SOLUTION: landscaping jobs. Since the landscaper will do at most Substitute –6 for x in f (x). 3 landscaping jobs per day, L ≤ 15. Since he works 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100.

If the landscaper does 3 landscaping jobs per day, he ANSWER: can only do 11 mowing jobs per day and work a total –21 of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. 68. g(–8) To maximize the earnings, write a function that SOLUTION: relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and Substitute –8 for x in g(x). $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of jobs. ANSWER: (M, L) F(M, L) = 35M + F(M, L) –64 125L 69. h(3) (100, 0) F(M, L) = 35(100) + 3500 SOLUTION: 125(0) Substitute 3 for x in h(x).

(55, 15) F(M, L) = 35(55) + 125 3800 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 (15)

15 landscape jobs and 55 lawns; $3800 ANSWER: –20 ANSWER: 15 landscape jobs and 55 lawns; $3800 70. f (c)

2 SOLUTION: Find each value if f (x) = 4x + 3, g(x) = –x , and Substitute c for x in f (x). 2 h(x) = –2x – 2x + 4. 67. f (–6) SOLUTION: Substitute –6 for x in f (x). ANSWER:

4c + 3

71. g(3d) SOLUTION: ANSWER: Substitute 3d for x in g(x). –21

68. g(–8) SOLUTION: ANSWER: Substitute –8 for x in g(x). 2 –9d

72. h(2b + 1) SOLUTION: ANSWER: Substitute 2b + 1 for x in h(x).

–64 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –8b2 – 12b

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

Simplify.

1. ANSWER:

SOLUTION:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) 4 3 2 SOLUTION: 5. (3z – 6z – 9z + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

ANSWER: A.

C. 5 2 6. (y – 3y – 20) ÷ (y – 2)

SOLUTION: D.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

The correct choice is A. ANSWER: ANSWER: A y 4 + 2y3 + 4y2 + 5y + 10 Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) 7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ? SOLUTION:

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) ANSWER: SOLUTION:

10.

SOLUTION:

ANSWER:

ANSWER: 2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

11.

SOLUTION:

ANSWER:

10. ANSWER: 3y + 5 SOLUTION: Simplify

12.

SOLUTION:

ANSWER: 3a2b – 2ab2 ANSWER:

13.

SOLUTION: 11.

SOLUTION:

ANSWER: x + 3y –2

14.

SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION: ANSWER:

ANSWER: 15. 3a2b – 2ab2 SOLUTION:

13.

SOLUTION:

ANSWER: 2 2a + b – 3

ANSWER: 16. x + 3y –2 SOLUTION:

14.

SOLUTION:

ANSWER: 4c2d2 – 6

ANSWER: 17.

SOLUTION: 15.

SOLUTION:

ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste ANSWER: that is reduced each day in a certain community can 2 2 2a + b – 3 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. 16. SOLUTION: SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: –b + 8 ANSWER: 19. BAKING The number of cookies produced in a 2 2 4c d – 6 factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by 17. w to find the average number of cookies produced per worker. SOLUTION: SOLUTION: The average number of cookies produced per worker is:

ANSWER: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can Simplify. 2 2 be estimated by –b + 8b, where b is the number of 20. (a – 8a – 26) ÷ (a + 2) bulbs. Divide by b to find the average amount of SOLUTION: energy saved per CFL bulb.

SOLUTION: The average amount of energy saved per CFL bulb is:

ANSWER: ANSWER: –b + 8

19. BAKING The number of cookies produced in a 21. (b3 – 4b2 + b – 2) ÷ (b + 1) factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by SOLUTION: w to find the average number of cookies produced per worker. SOLUTION: The average number of cookies produced per worker is:

ANSWER:

ANSWER: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1)

SOLUTION:

Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION:

ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) SOLUTION: ANSWER:

24.

ANSWER: SOLUTION:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) ANSWER: z3 – 2z2 – 4 SOLUTION:

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

ANSWER: 2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

24.

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) ANSWER:

SOLUTION:

27.

SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

ANSWER: 28.

SOLUTION:

27.

SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division: ANSWER:

28.

SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

ANSWER: Synthetic division:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2)

SOLUTION:

ANSWER:

Synthetic division: 6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

ANSWER:

Synthetic division:

SOLUTION:

The depressed polynomial is .

Since the volume of the box is the product of length, width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 2x + 3, 3x + 5 ANSWER: 33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a 32. CCSS REASONING A rectangular box for a new generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If product is designed in such a way that the three the current of the generator is I = t + 4, write an dimensions always have a particular relationship expression that represents the voltage. defined by the variable x. The volume of the box can 3 2 be written as 6x + 31x + 53x + 30, and the height is SOLUTION: always x + 2. What are the width and length of the box? SOLUTION: Divide the function by the height (x + 2) to find the Synthetic division: length and width. Use synthetic division.

The depressed polynomial is . ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer. Since the volume of the box is the product of length, • Choose a number and multiply it by 4. width and height, the width and length of box are (2x • Then add the sum of your number and 15 to the + 3)(3x + 5). product you found.

• Now divide by the sum of your number and 3. ANSWER: a. What number will the volunteer always have at the end? 2x + 3, 3x + 5 b. Explain the process you used to discover the 33. PHYSICS The voltage V is related to current I and answer. SOLUTION: power P by the equation . The power of a a. Let x be the number. generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient Synthetic division: is 5.

ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: 35. BUSINESS The number of magazine subscriptions 2 V(t) = t + 5t + 6 sold can be estimated by ,where a is 34. ENTERTAINMENT A magician gives these the amount of money the company spent on instructions to a volunteer. advertising in hundreds of dollars and n is the number • Choose a number and multiply it by 4. of subscriptions sold. • Then add the sum of your number and 15 to the product you found. a. Perform the division indicated by . • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the b. About how many subscriptions will be sold if end? $1500 is spent on advertising? b. Explain the process you used to discover the SOLUTION: answer.

SOLUTION: a. a. Let x be the number.

b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get b. There are 15 hundreds in 1500. Substitute a = 15. 5x + 15. Divide the polynomial by x + 3. The quotient is 5. ANSWER: a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions Therefore, about 2423 subscriptions will be sold. sold can be estimated by ,where a is ANSWER: the amount of money the company spent on advertising in hundreds of dollars and n is the number a. of subscriptions sold.

b. about 2423 subscriptions a. Perform the division indicated by . Simplify. b. About how many subscriptions will be sold if 36. (x4 – y4) ÷ (x – y) $1500 is spent on advertising? SOLUTION: SOLUTION:

ANSWER: 2 2 (x + y )(x + y)

3 2 2 b. There are 15 hundreds in 1500. Substitute a = 15. 37. (28c d – 21cd ) ÷ (14cd) SOLUTION:

ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) Therefore, about 2423 subscriptions will be sold. SOLUTION: ANSWER:

a. b. about 2423 subscriptions

Simplify. ANSWER: 36. (x4 – y4) ÷ (x – y) SOLUTION:

39.

SOLUTION:

ANSWER: 2 2 (x + y )(x + y)

37. (28c3d2 – 21cd2) ÷ (14cd) ANSWER: SOLUTION: n2 – n – 1

40.

ANSWER: SOLUTION:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER:

41.

SOLUTION: 39.

SOLUTION:

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9– ANSWER: 2 n – n – 1 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. 40. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. SOLUTION: b. SYMBOLIC Write an expression to represent

the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

ANSWER:

41.

SOLUTION:

The width is x + 3.

ANSWER: c.

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this Yes, the concrete model checks with the algebraic situation. Use the model to find the width. model. b. SYMBOLIC Write an expression to represent the model. ANSWER: c. NUMERICAL Solve this problem algebraically a. using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

The width is x + 3.

2 b. 2x + 7x + 3 ÷ (2x + 1)

The width is x + 3.

b. c.

yes c. 43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your reasoning. Yes, the concrete model checks with the algebraic model. SOLUTION: Use synthetic division. ANSWER: a.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a The width is x + 3. binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? b. 2x 2 + 7x + 3 ÷ (2x + 1) SOLUTION: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the c. binomial and the quotient.

ANSWER: yes The binomial is a factor of the polynomial.

43. ERROR ANALYSIS Sharon and Jamal are dividing 45. REASONING Review any of the division problems 3 2 in this lesson. What is the relationship between the 2x – 4x + 3x – 1 by x – 3. Sharon claims that the degrees of the dividend, the divisor, and the quotient? remainder is 26. Jamal argues that the remainder is – 100. Is either of them correct? Explain your SOLUTION: reasoning. Sample answer: For example:

SOLUTION:

Use synthetic division.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3.

ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3. In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. 44. CHALLENGE If a polynomial is divided by a The degree of the quotient plus the degree of the binomial and the remainder is 0, what does this tell divisor equals the degree of the dividend. you about the relationship between the binomial and the polynomial? ANSWER: Sample answer: The degree of the quotient plus the SOLUTION: degree of the divisor equals the degree of the If a polynomial divided by a binomial has no dividend. remainder, then the polynomial has two factors: the binomial and the quotient. 46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. ANSWER: The binomial is a factor of the polynomial. SOLUTION: Sample answer: Begin by multiplying two binomials 45. REASONING Review any of the division problems 2 such as (x + 2)(x + 3) which simplifies to x + 5x + in this lesson. What is the relationship between the 6. In order to get a remainder of 3 when divided, add degrees of the dividend, the divisor, and the quotient? 2 3 to the trinomial to get x + 5x + 9. When divided, SOLUTION: there will be a remainder of 3. The quotient of two Sample answer: For example: polynomials is .

ANSWER:

Sample answer:

47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your reasoning.

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the SOLUTION: divisor equals the degree of the dividend. does not belong with the other three. The other ANSWER: three expressions are polynomials. Since the Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the denominator of contains a variable, it is not a dividend. polynomial. 46. OPEN ENDED Write a quotient of two polynomials ANSWER: for which the remainder is 3. SOLUTION: does not belong with the other three. The other Sample answer: Begin by multiplying two binomials three expressions are polynomials. Since the such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add denominator of contains a variable, it is not a 2 3 to the trinomial to get x + 5x + 9. When divided, polynomial. there will be a remainder of 3. The quotient of two 48. WRITING IN MATH Use the information at the polynomials is . beginning of the lesson to write assembly instruction using the division of polynomials to make a paper ANSWER: cover for your textbook. SOLUTION: Sample answer: Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the 47. CCSS ARGUMENTS Identify the expression that quotient of 14x + 6 provides the length of the book does not belong with the other three. Explain your jacket. The front and back cover are each 6x units reasoning. long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is

the width of each flap. ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the SOLUTION: quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 does not belong with the other three. The other inches. Half of this length is the width of each flap. three expressions are polynomials. Since the 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number denominator of contains a variable, it is not a of women?

polynomial. A ANSWER: B does not belong with the other three. The other

C three expressions are polynomials. Since the

denominator of contains a variable, it is not a D polynomial. SOLUTION: 48. WRITING IN MATH Use the information at the Total number of employees: x + 3 beginning of the lesson to write assembly instruction Number of women: x using the division of polynomials to make a paper Ratio of the total number of employees to the number cover for your textbook. of women is:

SOLUTION: Sample answer: By dividing the area of the paper The correct choice is A. 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book ANSWER: jacket. The front and back cover are each 6x units A long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is 50. SAT/ACT Which polynomial has degree 3? the width of each flap. A x3 + x2 –2x4 2 ANSWER: B –2x – 3x + 4 2 C 3x – 3 Sample answer: By dividing 140x + 60x by 10x, the 2 3 quotient of 14x + 6 provides the length of the book D x + x + 12 3 jacket. Then, subtracting 14x, we are left with 6 E 1 + x + x inches. Half of this length is the width of each flap. SOLUTION: 49. An office employs x women and 3 men. What is the To determine the degree of a polynomial, look at the ratio of the total number of employees to the number term with the greatest exponent. of women? The correct choice is E.

A ANSWER: E B GRIDDED RESPONSE 51. In the figure below, C m + n + p = ?

SOLUTION: Total number of employees: x + 3 Number of women: x Ratio of the total number of employees to the number SOLUTION: of women is: Sum of the exterior angles of a triangle is . So,

The correct choice is A. ANSWER: 360 ANSWER: 2 2 A 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 2 F 2x 50. SAT/ACT Which polynomial has degree 3? 2 3 2 4 H –10x + 17x A x + x –2x 2 2 G –10x B –2x – 3x + 4 J 2 C 3x – 3 2x + 17x D x2 + x + 123 SOLUTION: 3 E 1 + x + x SOLUTION: To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is H. The correct choice is E. ANSWER: ANSWER: H E Simplify. 3 2 3 51. GRIDDED RESPONSE In the figure below, 53. (5x + 2x – 3x + 4) – (2x – 4x) m + n + p = ? SOLUTION: Use the Distributive Property and then combine like terms.

SOLUTION: Sum of the exterior angles of a triangle is . So, ANSWER: ANSWER: 3x3 + 2x2 + x + 4 360

2 2 3 2 52. (–4x + 2x + 3) – 3(2x – 5x + 1) = 54. (2y – 3y + 8) + (3y – 6y) F 2 2x SOLUTION: H –10x2 + 17x 2 G –10x J 2x2 + 17x ANSWER: SOLUTION: 3 2 2y + 3y – 9y + 8

55. 4a(2a – 3) + 3a(5a – 4) SOLUTION:

Use the Distributive Property and then combine like The correct choice is H. terms. ANSWER: H Simplify. ANSWER: 3 2 3 2 53. (5x + 2x – 3x + 4) – (2x – 4x) 23a – 24a SOLUTION: 56. (c + d)(c – d)(2c – 3d) Use the Distributive Property and then combine like SOLUTION: terms. Use the FOIL method and then combine like terms.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3

ANSWER: 2 2 3 57. (xy) (2xy z) 3x3 + 2x2 + x + 4 SOLUTION: Apply the properties of exponents to simplify the 3 2 54. (2y – 3y + 8) + (3y – 6y) expression.

SOLUTION:

ANSWER: ANSWER: 8x5y 8z 3 2y 3 + 3y 2 – 9y + 8 2 –2 2 2 58. (3ab ) (2a b) 55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: SOLUTION: Use the Distributive Property and then combine like terms.

ANSWER: 2 ANSWER: 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: LANDSCAPING Use the FOIL method and then combine like terms. 59. Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to ANSWER: cover no more than 240 square feet. What could the width of his garden be? 2c3 – 3c2d – 2cd 2 + 3d 3 SOLUTION: 2 2 3 57. (xy) (2xy z) Let x be the length and y be the width of the garden.

SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: 8x5y 8z 3 The solution of the inequality is . Since y represent the 2 –2 2 2 58. (3ab ) (2a b) width of the garden and the sum of the length and SOLUTION: width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 ANSWER: SOLUTION:

SOLUTION: Let x be the length and y be the width of the garden. ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 5-2 DividingANSWER: Polynomials 0 to 10 ft or 24 to 34 ft

Solve each equation by completing the square. 62. 2x2 + 6x + 5 = 0 60. x2 + 6x + 2 = 0 SOLUTION: SOLUTION:

ANSWER: ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 State the consecutive integers between which the zeros of each quadratic function are located. SOLUTION:

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER: ANSWER: between –6 and –5; between –3 and –2

62. 2x2 + 6x + 5 = 0

SOLUTION: 64. SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3

65. ANSWER: SOLUTION:

The sign of f (x) changes between the x values eSolutions Manual - Powered by Cognero between –1 and 0, and between 2 and 3. So thePage 15 zeros of the quadratic function lies between –1 and 0, State the consecutive integers between which and between 2 and 3. the zeros of each quadratic function are located. ANSWER: 63. between –1 and 0; between 2 and 3

SOLUTION: 66. BUSINESS A landscaper can mow a lawn in 30 The sign of f (x) changes between the x values –6 minutes and perform a small landscape job in 90 and –5, and between –3 and –2. So the zeros of the minutes. He works at most 10 hours per day, 5 days quadratic function lies between –6 and –5, and per week. He earns $35 per lawn and $125 per between –3 and –2. landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed ANSWER: and completed landscape jobs per week that will between –6 and –5; between –3 and –2 maximize income. Then find the maximum income. SOLUTION: Let M be the lawns mowed and L be the small 64. landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works SOLUTION: 10 hours per day, 5 days a week, he can do at most The sign of f (x) changes between the x values 20 mowing jobs per day or 100 per week. So, M ≤ between 0 and 1, and between 2 and 3. So the zeros 100. of the quadratic function lies between 0 and 1, and between 2 and 3. If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total ANSWER: of 10 hours per day. So, for the week, if he does 15 between 0 and 1; between 2 and 3 landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that relates the number and charge for each type of job. 65. Since he charges $35 for each lawn mowed and SOLUTION: $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and The sign of f (x) changes between the x values L to determine the earnings for each combination of between –1 and 0, and between 2 and 3. So the jobs. zeros of the quadratic function lies between –1 and 0, and between 2 and 3. (M, L) F(M, L) = 35M + F(M, L) ANSWER: 125L between –1 and 0; between 2 and 3 (100, 0) F(M, L) = 35(100) + 3500 66. BUSINESS A landscaper can mow a lawn in 30 125(0) minutes and perform a small landscape job in 90 minutes. He works at most 10 hours per day, 5 days (55, 15) F(M, L) = 35(55) + 125 3800 per week. He earns $35 per lawn and $125 per (15) landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed (0, 15) F(M, L) = 35(0) + 125 1875 and completed landscape jobs per week that will (15) maximize income. Then find the maximum income.

SOLUTION: 15 landscape jobs and 55 lawns; $3800 Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most ANSWER: 3 landscaping jobs per day, L ≤ 15. Since he works 15 landscape jobs and 55 lawns; $3800 10 hours per day, 5 days a week, he can do at most 2 20 mowing jobs per day or 100 per week. So, M ≤ Find each value if f (x) = 4x + 3, g(x) = –x , and 100. h(x) = –2x2 – 2x + 4. 67. f (–6) If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total SOLUTION: of 10 hours per day. So, for the week, if he does 15 Substitute –6 for x in f (x). landscaping jobs, he can mow 55 lawns.

To maximize the earnings, write a function that relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and ANSWER: L to determine the earnings for each combination of –21 jobs. 68. g(–8) (M, L) F(M, L) = 35M + F(M, L) SOLUTION: 125L Substitute –8 for x in g(x).

(100, 0) F(M, L) = 35(100) + 3500 125(0)

(55, 15) F(M, L) = 35(55) + 125 3800 (15) ANSWER: –64 (0, 15) F(M, L) = 35(0) + 125 1875 (15) 69. h(3) SOLUTION: 15 landscape jobs and 55 lawns; $3800 Substitute 3 for x in h(x).

ANSWER: 15 landscape jobs and 55 lawns; $3800

Find each value if f (x) = 4x + 3, g(x) = –x2, and h(x) = –2x2 – 2x + 4. 67. f (–6) ANSWER: SOLUTION: –20 Substitute –6 for x in f (x). 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: –21 ANSWER: 68. g(–8) 4c + 3 SOLUTION: 71. g(3d) Substitute –8 for x in g(x). SOLUTION:

Substitute 3d for x in g(x).

ANSWER: –64 ANSWER: 2 69. h(3) –9d SOLUTION: 72. h(2b + 1) Substitute 3 for x in h(x). SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –20 70. f (c) ANSWER: –8b2 – 12b SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) SOLUTION: Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

Simplify. 2 2 –1 2. (3a b – 6ab + 5ab )(ab) 1. SOLUTION:

SOLUTION:

ANSWER: 3a + 5b – 6 ANSWER: 4y + 2x – 2 3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION: 2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) ANSWER: SOLUTION:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION: ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

ANSWER: D. 4 3 2 y + 2y + 4y + 5y + 10 SOLUTION: 7. MULTIPLE CHOICE Which expression is equal First rewrite the divisor so the x-term is first. Then

2 –1 use long division. to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION:

First rewrite the divisor so the x-term is first. Then use long division.

The correct choice is A. ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5)

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: ANSWER:

10.

SOLUTION:

ANSWER:

10.

SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

11.

SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: ANSWER: 3a2b – 2ab2 3y + 5

Simplify 13.

12. SOLUTION: SOLUTION:

ANSWER: 3a2b – 2ab2

ANSWER:

13. x + 3y –2

SOLUTION: 14.

SOLUTION:

ANSWER: x + 3y –2 ANSWER:

14.

SOLUTION: 15.

SOLUTION:

ANSWER:

ANSWER: 15. 2 2a + b – 3

SOLUTION: 16.

SOLUTION:

ANSWER: 2 2a + b – 3 ANSWER:

16. 4c2d2 – 6

SOLUTION: 17.

SOLUTION:

ANSWER: 2 2 4c d – 6 ANSWER: 3np – 6 + 7p

17. 18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste SOLUTION: that is reduced each day in a certain community can 2 be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb ANSWER: is: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs reduce energy waste. The amount of energy waste that is reduced each day in a certain community can ANSWER: 2 be estimated by –b + 8b, where b is the number of –b + 8 bulbs. Divide by b to find the average amount of energy saved per CFL bulb. 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + SOLUTION: 1000, where w is the number of workers. Divide by The average amount of energy saved per CFL bulb w to find the average number of cookies produced is: per worker.

SOLUTION: The average number of cookies produced per worker is: ANSWER: –b + 8

19. BAKING The number of cookies produced in a 2 factory each day can be estimated by –w + 16w + ANSWER: 1000, where w is the number of workers. Divide by

w to find the average number of cookies produced per worker. SOLUTION: Simplify. 2 The average number of cookies produced per worker 20. (a – 8a – 26) ÷ (a + 2) is: SOLUTION:

ANSWER:

ANSWER: Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) SOLUTION: 3 2 21. (b – 4b + b – 2) ÷ (b + 1) SOLUTION:

ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) 4 3 2 –1 SOLUTION: 22. (z – 3z + 2z – 4z + 4)(z – 1) SOLUTION:

ANSWER: ANSWER: z3 – 2z2 – 4

4 3 2 –1 5 3 2 22. (z – 3z + 2z – 4z + 4)(z – 1) 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION: SOLUTION:

ANSWER: ANSWER: 3 2 z – 2z – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4)

SOLUTION: 24.

SOLUTION:

ANSWER:

24.

SOLUTION: 4 2 25. (g – 3g – 18) ÷ (g – 2) SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) 2 SOLUTION: 26. (6a – 3a + 9) ÷ (3a – 2)

SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION:

ANSWER:

27.

SOLUTION:

ANSWER:

27.

SOLUTION:

ANSWER:

28.

SOLUTION:

ANSWER:

28.

ANSWER: SOLUTION:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

ANSWER: Synthetic division:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

ANSWER:

Synthetic division:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: ANSWER:

Synthetic division: 6 5 3 −1 31. (10y + 5y + 10y – 20y – 15)(5y + 5) SOLUTION:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1

SOLUTION:

ANSWER:

always x + 2. What are the width and length of the The depressed polynomial is . box?

SOLUTION: Divide the function by the height (x + 2) to find the length and width. Use synthetic division. Since the volume of the box is the product of length,

width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER:

2x + 3, 3x + 5 The depressed polynomial is . 33. PHYSICS The voltage V is related to current I and

power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If Since the volume of the box is the product of length, the current of the generator is I = t + 4, write an width and height, the width and length of box are (2x expression that represents the voltage. + 3)(3x + 5). SOLUTION:

ANSWER: 2x + 3, 3x + 5 Synthetic division: 33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage.

SOLUTION: ANSWER: 2 V(t) = t + 5t + 6

Synthetic division: 34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the ANSWER: answer. 2 V(t) = t + 5t + 6 SOLUTION: a. Let x be the number. 34. ENTERTAINMENT A magician gives these

instructions to a volunteer. • Choose a number and multiply it by 4. • Then add the sum of your number and 15 to the product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the b. Sample answer: Let x be the number. Multiply the end? x by 4 to get 4x. Then add x +15 to the product to get b. Explain the process you used to discover the 5x + 15. Divide the polynomial by x + 3. The quotient answer. is 5.

SOLUTION: ANSWER: a. Let x be the number. a.5 b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get sold can be estimated by ,where a is 5x + 15. Divide the polynomial by x + 3. The quotient is 5. the amount of money the company spent on advertising in hundreds of dollars and n is the number ANSWER: of subscriptions sold. a.5 b. Sample answer: Let x be the number. a. Perform the division indicated by . Multiply the x by 4 to get 4x. Then add x + 15 to the b. product to get 5x + 15. Divide the polynomial by x + About how many subscriptions will be sold if

3. The quotient is 5. $1500 is spent on advertising? SOLUTION: BUSINESS 35. The number of magazine subscriptions sold can be estimated by ,where a is a. the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? b. There are 15 hundreds in 1500. Substitute a = 15. SOLUTION:

Therefore, about 2423 subscriptions will be sold. b. There are 15 hundreds in 1500. Substitute a = 15. ANSWER:

a. b. about 2423 subscriptions

Simplify. 36. (x4 – y4) ÷ (x – y) SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER:

a. b. about 2423 subscriptions ANSWER: 2 2 (x + y )(x + y) Simplify. 36. (x4 – y4) ÷ (x – y) 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER: ANSWER: 2 2 (x + y )(x + y)

3 2 2 –1 38. (a b – a b + 2b)(–ab) 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION: SOLUTION:

ANSWER: ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) 39. SOLUTION: SOLUTION:

ANSWER:

ANSWER: n2 – n – 1 39.

SOLUTION: 40.

SOLUTION:

ANSWER: n2 – n – 1 ANSWER:

40.

SOLUTION:

41.

SOLUTION:

ANSWER:

41. 3z 4 – z 3 + 2 z2 – 4z + 9– SOLUTION: 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete ANSWER: model check with your algebraic model? 4 3 2 3z – z + 2 z – 4z + 9– SOLUTION: a. 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent the model. c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION:

a. The width is x + 3.

The width is x + 3. Yes, the concrete model checks with the algebraic model.

b. ANSWER: a.

Yes, the concrete model checks with the algebraic model.

ANSWER:

a. The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing The width is x + 3. 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the 2 remainder is 26. Jamal argues that the remainder is – b. 2x + 7x + 3 ÷ (2x + 1) 100. Is either of them correct? Explain your reasoning.

SOLUTION: c. Use synthetic division.

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the The remainder is 26. So, Sharon is correct. Jamal remainder is 26. Jamal argues that the remainder is – actually divided by x + 3. 100. Is either of them correct? Explain your reasoning. ANSWER: SOLUTION: Sample answer: Sharon; Jamal actually divided by x Use synthetic division. + 3.

44. CHALLENGE If a polynomial is divided by a binomial and the remainder is 0, what does this tell you about the relationship between the binomial and the polynomial? SOLUTION: The remainder is 26. So, Sharon is correct. Jamal If a polynomial divided by a binomial has no actually divided by x + 3. remainder, then the polynomial has two factors: the binomial and the quotient. ANSWER: Sample answer: Sharon; Jamal actually divided by x ANSWER: + 3. The binomial is a factor of the polynomial.

44. CHALLENGE If a polynomial is divided by a 45. REASONING Review any of the division problems binomial and the remainder is 0, what does this tell in this lesson. What is the relationship between the you about the relationship between the binomial and degrees of the dividend, the divisor, and the quotient? the polynomial? SOLUTION: SOLUTION: Sample answer: For example: If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example:

In this exercise, the degree of the dividend is 2. The degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

ANSWER: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3.

In this exercise, the degree of the dividend is 2. The SOLUTION: degree of the divisor and the quotient are each 1. Sample answer: Begin by multiplying two binomials The degree of the quotient plus the degree of the such as (x + 2)(x + 3) which simplifies to x2 + 5x +

divisor equals the degree of the dividend. 6. In order to get a remainder of 3 when divided, add 2 ANSWER: 3 to the trinomial to get x + 5x + 9. When divided, Sample answer: The degree of the quotient plus the there will be a remainder of 3. The quotient of two degree of the divisor equals the degree of the dividend. polynomials is .

46. OPEN ENDED Write a quotient of two polynomials ANSWER: for which the remainder is 3. Sample answer: SOLUTION: Sample answer: Begin by multiplying two binomials 2 47. CCSS ARGUMENTS Identify the expression that such as (x + 2)(x + 3) which simplifies to x + 5x + does not belong with the other three. Explain your 6. In order to get a remainder of 3 when divided, add 2 reasoning. 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two

polynomials is .

ANSWER: SOLUTION: Sample answer: does not belong with the other three. The other 47. CCSS ARGUMENTS Identify the expression that three expressions are polynomials. Since the does not belong with the other three. Explain your reasoning. denominator of contains a variable, it is not a polynomial.

ANSWER:

does not belong with the other three. The other

SOLUTION: three expressions are polynomials. Since the denominator of contains a variable, it is not a does not belong with the other three. The other polynomial. three expressions are polynomials. Since the denominator of contains a variable, it is not a 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction polynomial. using the division of polynomials to make a paper cover for your textbook. ANSWER: SOLUTION: does not belong with the other three. The other Sample answer: By dividing the area of the paper 2 three expressions are polynomials. Since the 140x + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book denominator of contains a variable, it is not a jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting polynomial. 14x, we are left with 6 inches. Half of this length is the width of each flap. 48. WRITING IN MATH Use the information at the beginning of the lesson to write assembly instruction ANSWER: using the division of polynomials to make a paper 2 cover for your textbook. Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book SOLUTION: jacket. Then, subtracting 14x, we are left with 6 Sample answer: By dividing the area of the paper inches. Half of this length is the width of each flap. 140x2 + 60x by the height of the book jacket 10x, the quotient of 14x + 6 provides the length of the book 49. An office employs x women and 3 men. What is the jacket. The front and back cover are each 6x units ratio of the total number of employees to the number

long and the spine is 2x units long. Then, subtracting of women?

14x, we are left with 6 inches. Half of this length is A the width of each flap.

ANSWER: B 2 Sample answer: By dividing 140x + 60x by 10x, the C quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. D

49. An office employs x women and 3 men. What is the SOLUTION: ratio of the total number of employees to the number Total number of employees: x + 3

of women? Number of women: x

A Ratio of the total number of employees to the number of women is:

B The correct choice is A. C ANSWER: D A

SOLUTION: 50. SAT/ACT Which polynomial has degree 3? 3 2 4 Total number of employees: x + 3 A x + x –2x 2 Number of women: x B –2x – 3x + 4 Ratio of the total number of employees to the number C 3x – 3 of women is: 2 3 D x + x + 12 3 E 1 + x + x The correct choice is A. SOLUTION: ANSWER: To determine the degree of a polynomial, look at the A term with the greatest exponent. The correct choice is E. 50. SAT/ACT Which polynomial has degree 3? ANSWER: A x3 + x2 –2x4 2 E B –2x – 3x + 4 C 3x – 3 51. GRIDDED RESPONSE In the figure below, D x2 + x + 123 m + n + p = ? 3 E 1 + x + x SOLUTION: To determine the degree of a polynomial, look at the

term with the greatest exponent. The correct choice is E. SOLUTION: ANSWER: Sum of the exterior angles of a triangle is . E So,

51. GRIDDED RESPONSE In the figure below, ANSWER: m + n + p = ? 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x H –10x2 + 17x 2 G –10x SOLUTION: J 2x2 + 17x Sum of the exterior angles of a triangle is . SOLUTION: So,

ANSWER: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = The correct choice is H. 2 F 2x ANSWER: 2 H –10x + 17x H 2 G –10x 2 Simplify. J 2x + 17x 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: SOLUTION: Use the Distributive Property and then combine like terms.

The correct choice is H.

ANSWER: H

Simplify. ANSWER: 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) 3x3 + 2x2 + x + 4 SOLUTION: Use the Distributive Property and then combine like 3 2 54. (2y – 3y + 8) + (3y – 6y) terms. SOLUTION:

ANSWER: 2y 3 + 3y 2 – 9y + 8 ANSWER: 55. 4a(2a – 3) + 3a(5a – 4) 3x3 + 2x2 + x + 4 SOLUTION: 3 2 Use the Distributive Property and then combine like 54. (2y – 3y + 8) + (3y – 6y) terms. SOLUTION:

ANSWER: 2 ANSWER: 23a – 24a 3 2 2y + 3y – 9y + 8 56. (c + d)(c – d)(2c – 3d) 55. 4a(2a – 3) + 3a(5a – 4) SOLUTION: SOLUTION: Use the FOIL method and then combine like terms. Use the Distributive Property and then combine like terms.

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3 ANSWER: 2 2 2 3 23a – 24a 57. (xy) (2xy z) SOLUTION: 56. (c + d)(c – d)(2c – 3d) Apply the properties of exponents to simplify the SOLUTION: expression. Use the FOIL method and then combine like terms.

ANSWER: ANSWER: 8x5y 8z 3 2c3 – 3c2d – 2cd 2 + 3d 3 2 –2 2 2 2 2 3 58. (3ab ) (2a b) 57. (xy) (2xy z) SOLUTION: SOLUTION: Apply the properties of exponents to simplify the expression.

ANSWER: ANSWER: 5 8 3 8x y z 2 –2 2 2 58. (3ab ) (2a b) SOLUTION: 59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: ANSWER: Let x be the length and y be the width of the garden.

59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? The solution of the inequality is SOLUTION: . Since y represent the Let x be the length and y be the width of the garden. width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft Solve each equation by completing the square. 60. x2 + 6x + 2 = 0

The solution of the inequality is SOLUTION: . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft

Solve each equation by completing the square. 2 60. x + 6x + 2 = 0

SOLUTION: ANSWER:

–3 ±

2 61. x – 8x – 3 = 0 SOLUTION:

ANSWER:

–3 ± ANSWER: 2 61. x – 8x – 3 = 0 SOLUTION: 62. 2x2 + 6x + 5 = 0 SOLUTION:

ANSWER:

62. 2x2 + 6x + 5 = 0 SOLUTION: ANSWER:

State the consecutive integers between which the zeros of each quadratic function are located.

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the ANSWER: quadratic function lies between –6 and –5, and between –3 and –2. ANSWER: State the consecutive integers between which between –6 and –5; between –3 and –2 the zeros of each quadratic function are located.

63. 64. SOLUTION: SOLUTION: The sign of f (x) changes between the x values –6 The sign of f (x) changes between the x values and –5, and between –3 and –2. So the zeros of the between 0 and 1, and between 2 and 3. So the zeros quadratic function lies between –6 and –5, and of the quadratic function lies between 0 and 1, and between –3 and –2. between 2 and 3.

ANSWER: ANSWER: between –6 and –5; between –3 and –2 between 0 and 1; between 2 and 3

64. 65. SOLUTION: SOLUTION: The sign of f (x) changes between the x values The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros between –1 and 0, and between 2 and 3. So the of the quadratic function lies between 0 and 1, and zeros of the quadratic function lies between –1 and 0, between 2 and 3. and between 2 and 3.

5-2 DividingANSWER: Polynomials ANSWER: between 0 and 1; between 2 and 3 between –1 and 0; between 2 and 3

66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 65. minutes. He works at most 10 hours per day, 5 days per week. He earns $35 per lawn and $125 per SOLUTION: landscape job. He cannot do more than 3 landscape The sign of f (x) changes between the x values jobs per day. Find the combination of lawns mowed between –1 and 0, and between 2 and 3. So the and completed landscape jobs per week that will zeros of the quadratic function lies between –1 and 0, maximize income. Then find the maximum income. and between 2 and 3. SOLUTION: ANSWER: Let M be the lawns mowed and L be the small between –1 and 0; between 2 and 3 landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 66. BUSINESS A landscaper can mow a lawn in 30 10 hours per day, 5 days a week, he can do at most minutes and perform a small landscape job in 90 20 mowing jobs per day or 100 per week. So, M ≤ minutes. He works at most 10 hours per day, 5 days 100. per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape If the landscaper does 3 landscaping jobs per day, he jobs per day. Find the combination of lawns mowed can only do 11 mowing jobs per day and work a total and completed landscape jobs per week that will of 10 hours per day. So, for the week, if he does 15 maximize income. Then find the maximum income. landscaping jobs, he can mow 55 lawns. SOLUTION: To maximize the earnings, write a function that Let M be the lawns mowed and L be the small relates the number and charge for each type of job. landscaping jobs. Since the landscaper will do at most Since he charges $35 for each lawn mowed and 3 landscaping jobs per day, L 15. Since he works ≤ $125 for each small landscaping job, F(M, L) = 35M 10 hours per day, 5 days a week, he can do at most + 125L. Next, substitute in the key values of M and 20 mowing jobs per day or 100 per week. So, M ≤ L to determine the earnings for each combination of 100. jobs.

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total (M, L) F(M, L) = 35M + F(M, L) of 10 hours per day. So, for the week, if he does 15 125L landscaping jobs, he can mow 55 lawns. (100, 0) F(M, L) = 35(100) + 3500 To maximize the earnings, write a function that 125(0) relates the number and charge for each type of job. Since he charges $35 for each lawn mowed and (55, 15) F(M, L) = 35(55) + 125 3800 $125 for each small landscaping job, F(M, L) = 35M (15) + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of (0, 15) F(M, L) = 35(0) + 125 1875 jobs. (15)

(M, L) F(M, L) = 35M + F(M, L) 15 landscape jobs and 55 lawns; $3800 125L ANSWER: (100, 0) F(M, L) = 35(100) + 3500 15 landscape jobs and 55 lawns; $3800 125(0) Find each value if f (x) = 4x + 3, g(x) = –x2, and (55, 15) F(M, L) = 35(55) + 125 3800 h(x) = –2x2 – 2x + 4. (15) 67. f (–6) (0, 15) F(M, L) = 35(0) + 125 1875 SOLUTION: eSolutions Manual - Powered(15) by Cognero Substitute –6 for x in f (x). Page 16

15 landscape jobs and 55 lawns; $3800

ANSWER: 15 landscape jobs and 55 lawns; $3800 ANSWER: 2 Find each value if f (x) = 4x + 3, g(x) = –x , and –21 h(x) = –2x2 – 2x + 4. 67. f (–6) 68. g(–8) SOLUTION: SOLUTION: Substitute –6 for x in f (x). Substitute –8 for x in g(x).

ANSWER: ANSWER: –64 –21 69. h(3) 68. g(–8) SOLUTION: SOLUTION: Substitute 3 for x in h(x). Substitute –8 for x in g(x).

ANSWER: –64 ANSWER: –20 69. h(3) SOLUTION: 70. f (c) Substitute 3 for x in h(x). SOLUTION: Substitute c for x in f (x).

ANSWER: ANSWER: 4c + 3 –20 71. g(3d) 70. f (c) SOLUTION: SOLUTION: Substitute 3d for x in g(x).

Substitute c for x in f (x).

ANSWER: 2 ANSWER: –9d 4c + 3 72. h(2b + 1) 71. g(3d) SOLUTION: SOLUTION: Substitute 2b + 1 for x in h(x). Substitute 3d for x in g(x).

ANSWER: 2 –9d ANSWER: 72. h(2b + 1) –8b2 – 12b SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b Simplify.

SOLUTION:

ANSWER: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

Simplify.

1. ANSWER: SOLUTION:

5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3)

ANSWER: SOLUTION: 4y + 2x – 2

2 2 –1 2. (3a b – 6ab + 5ab )(ab) SOLUTION:

ANSWER: 3a + 5b – 6

3. (x2 – 6x – 20) ÷ (x + 2) SOLUTION:

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2) SOLUTION: ANSWER:

2 4. (2a – 4a – 8) ÷ (a + 1) SOLUTION:

ANSWER:

ANSWER: y 4 + 2y3 + 4y2 + 5y + 10 5. (3z4 – 6z3 – 9z2 + 3z – 6) ÷ (z + 3) MULTIPLE CHOICE SOLUTION: 7. Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. B.

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

5 2 6. (y – 3y – 20) ÷ (y – 2)

SOLUTION: The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION:

ANSWER: 4 3 2 y + 2y + 4y + 5y + 10 ANSWER:

7. MULTIPLE CHOICE Which expression is equal 2 –1 to (x + 3x – 9)(4 – x) ?

A. 2 9. (18a + 6a + 9) ÷ (3a – 2) B. SOLUTION:

SOLUTION: First rewrite the divisor so the x-term is first. Then use long division.

ANSWER:

10.

SOLUTION:

The correct choice is A.

ANSWER: A Simplify. 8. (10x2 + 15x + 20) ÷ (5x + 5) SOLUTION: ANSWER:

11.

SOLUTION:

ANSWER:

2 9. (18a + 6a + 9) ÷ (3a – 2) SOLUTION: ANSWER: 3y + 5 Simplify

12.

SOLUTION:

ANSWER: ANSWER: 2 2 3a b – 2ab

13. 10. SOLUTION: SOLUTION:

ANSWER: x + 3y –2

ANSWER: 14.

SOLUTION:

11.

SOLUTION:

ANSWER:

15.

SOLUTION:

ANSWER: 3y + 5 Simplify

12. ANSWER: SOLUTION: 2 2a + b – 3

16. ANSWER: 3a2b – 2ab2 SOLUTION:

13.

SOLUTION:

ANSWER: 4c2d2 – 6

17.

ANSWER: SOLUTION: x + 3y –2

14.

SOLUTION:

ANSWER: 3np – 6 + 7p

15. SOLUTION: The average amount of energy saved per CFL bulb SOLUTION: is:

ANSWER: –b + 8

19. BAKING The number of cookies produced in a ANSWER: factory each day can be estimated by –w2 + 16w + 2 1000, where w is the number of workers. Divide by 2a + b – 3 w to find the average number of cookies produced per worker. 16. SOLUTION: The average number of cookies produced per worker SOLUTION: is:

ANSWER:

ANSWER: 2 2 4c d – 6 Simplify. 2 20. (a – 8a – 26) ÷ (a + 2) 17. SOLUTION:

SOLUTION:

ANSWER: ANSWER: 3np – 6 + 7p

18. ENERGY Compact fluorescent light (CFL) bulbs 3 2 reduce energy waste. The amount of energy waste 21. (b – 4b + b – 2) ÷ (b + 1) that is reduced each day in a certain community can 2 SOLUTION: be estimated by –b + 8b, where b is the number of bulbs. Divide by b to find the average amount of energy saved per CFL bulb. SOLUTION: The average amount of energy saved per CFL bulb is: ANSWER:

ANSWER: 4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) –b + 8 SOLUTION: 19. BAKING The number of cookies produced in a factory each day can be estimated by –w2 + 16w + 1000, where w is the number of workers. Divide by w to find the average number of cookies produced per worker. SOLUTION: ANSWER: The average number of cookies produced per worker 3 2 is: z – 2z – 4 5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

Simplify. 2 ANSWER: 20. (a – 8a – 26) ÷ (a + 2)

SOLUTION:

24.

SOLUTION:

ANSWER:

21. (b3 – 4b2 + b – 2) ÷ (b + 1) ANSWER: SOLUTION:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

4 3 2 –1 22. (z – 3z + 2z – 4z + 4)(z – 1) ANSWER: SOLUTION:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION: ANSWER: z3 – 2z2 – 4

5 3 2 23. (x – 4x + 4x ) ÷ (x – 4) SOLUTION:

ANSWER:

24.

SOLUTION: 27.

SOLUTION:

ANSWER:

25. (g4 – 3g2 – 18) ÷ (g – 2) SOLUTION:

ANSWER:

2 26. (6a – 3a + 9) ÷ (3a – 2) SOLUTION: 28.

SOLUTION:

ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

27.

SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 30. (6z + 3z – 9z )(3z – 6) SOLUTION: ANSWER:

Synthetic division:

28.

SOLUTION:

ANSWER:

31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 SOLUTION: ANSWER:

29. (2b3 – 6b2 + 8b) ÷ (2b + 2) SOLUTION:

Synthetic division:

ANSWER:

6 4 2 –1 SOLUTION: 30. (6z + 3z – 9z )(3z – 6) Divide the function by the height (x + 2) to find the SOLUTION: length and width. Use synthetic division.

Synthetic division:

The depressed polynomial is .

Since the volume of the box is the product of length, ANSWER: width and height, the width and length of box are (2x + 3)(3x + 5).

ANSWER: 31. (10y6 + 5y5 + 10y3 – 20y – 15)(5y + 5)−1 2x + 3, 3x + 5 SOLUTION: 33. PHYSICS The voltage V is related to current I and power P by the equation . The power of a

generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an expression that represents the voltage. SOLUTION:

Synthetic division:

ANSWER:

32. CCSS REASONING A rectangular box for a new product is designed in such a way that the three ANSWER: dimensions always have a particular relationship 2 defined by the variable x. The volume of the box can V(t) = t + 5t + 6 3 2 be written as 6x + 31x + 53x + 30, and the height is 34. ENTERTAINMENT A magician gives these always x + 2. What are the width and length of the instructions to a volunteer. box? • Choose a number and multiply it by 4. SOLUTION: • Then add the sum of your number and 15 to the product you found. Divide the function by the height (x + 2) to find the • Now divide by the sum of your number and 3. length and width. Use synthetic division. a. What number will the volunteer always have at the

end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number. The depressed polynomial is .

Since the volume of the box is the product of length, b. Sample answer: Let x be the number. Multiply the width and height, the width and length of box are (2x x by 4 to get 4x. Then add x +15 to the product to get + 3)(3x + 5). 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

ANSWER: ANSWER: 2x + 3, 3x + 5 a.5 b. Sample answer: Let x be the number. 33. PHYSICS The voltage V is related to current I and Multiply the x by 4 to get 4x. Then add x + 15 to the power P by the equation . The power of a product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5. generator is modeled by P(t) = t3 + 9t2 + 26t + 24. If the current of the generator is I = t + 4, write an 35. BUSINESS The number of magazine subscriptions expression that represents the voltage. sold can be estimated by ,where a is SOLUTION: the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold.

Synthetic division: a. Perform the division indicated by .

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION:

ANSWER: 2 V(t) = t + 5t + 6

34. ENTERTAINMENT A magician gives these instructions to a volunteer. • Choose a number and multiply it by 4. b. • Then add the sum of your number and 15 to the There are 15 hundreds in 1500. Substitute a = 15. product you found. • Now divide by the sum of your number and 3. a. What number will the volunteer always have at the end? b. Explain the process you used to discover the answer. SOLUTION: a. Let x be the number.

Therefore, about 2423 subscriptions will be sold.

ANSWER:

a. b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x +15 to the product to get b. about 2423 subscriptions 5x + 15. Divide the polynomial by x + 3. The quotient is 5. Simplify. 36. (x4 – y4) ÷ (x – y) ANSWER: a.5 SOLUTION: b. Sample answer: Let x be the number. Multiply the x by 4 to get 4x. Then add x + 15 to the product to get 5x + 15. Divide the polynomial by x + 3. The quotient is 5.

35. BUSINESS The number of magazine subscriptions ANSWER: sold can be estimated by ,where a is 2 2 (x + y )(x + y) the amount of money the company spent on advertising in hundreds of dollars and n is the number of subscriptions sold. 37. (28c3d2 – 21cd2) ÷ (14cd)

a. Perform the division indicated by . SOLUTION:

b. About how many subscriptions will be sold if $1500 is spent on advertising? SOLUTION: ANSWER: a.

3 2 2 –1 38. (a b – a b + 2b)(–ab)

SOLUTION:

b. There are 15 hundreds in 1500. Substitute a = 15.

ANSWER:

39.

SOLUTION:

Therefore, about 2423 subscriptions will be sold.

ANSWER:

b. about 2423 subscriptions ANSWER: 2 Simplify. n – n – 1 4 4 36. (x – y ) ÷ (x – y) SOLUTION: 40.

SOLUTION:

ANSWER: 2 2 (x + y )(x + y) ANSWER: 37. (28c3d2 – 21cd2) ÷ (14cd) SOLUTION:

41.

SOLUTION: ANSWER:

3 2 2 –1 38. (a b – a b + 2b)(–ab) SOLUTION:

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

ANSWER: 42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this situation. Use the model to find the width.

39. b. SYMBOLIC Write an expression to represent the model. c NUMERICAL SOLUTION: . Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? SOLUTION: a.

ANSWER: n2 – n – 1

40.

SOLUTION:

The width is x + 3.

ANSWER: c.

41. Yes, the concrete model checks with the algebraic SOLUTION: model.

ANSWER: a.

ANSWER:

3z 4 – z 3 + 2 z2 – 4z + 9–

42. MULTIPLE REPRESENTATIONS Consider a 2 rectangle with area 2x + 7x + 3 and length 2x + 1. a. CONCRETE Use algebra tiles to represent this The width is x + 3. situation. Use the model to find the width. b. SYMBOLIC Write an expression to represent b. 2 the model. 2x + 7x + 3 ÷ (2x + 1) c. NUMERICAL Solve this problem algebraically using synthetic or long division. Does your concrete model check with your algebraic model? c. SOLUTION: a. yes

The width is x + 3.

The remainder is 26. So, Sharon is correct. Jamal actually divided by x + 3. c. ANSWER: Sample answer: Sharon; Jamal actually divided by x + 3.

44. CHALLENGE If a polynomial is divided by a Yes, the concrete model checks with the algebraic binomial and the remainder is 0, what does this tell model. you about the relationship between the binomial and the polynomial? ANSWER: SOLUTION: a. If a polynomial divided by a binomial has no remainder, then the polynomial has two factors: the binomial and the quotient.

ANSWER: The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient? SOLUTION: Sample answer: For example: The width is x + 3.

b. 2x 2 + 7x + 3 ÷ (2x + 1)

yes

43. ERROR ANALYSIS Sharon and Jamal are dividing 3 2 2x – 4x + 3x – 1 by x – 3. Sharon claims that the remainder is 26. Jamal argues that the remainder is – In this exercise, the degree of the dividend is 2. The 100. Is either of them correct? Explain your degree of the divisor and the quotient are each 1. reasoning. The degree of the quotient plus the degree of the divisor equals the degree of the dividend. SOLUTION: Use synthetic division. ANSWER: Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the dividend.

46. OPEN ENDED Write a quotient of two polynomials for which the remainder is 3. The remainder is 26. So, Sharon is correct. Jamal SOLUTION: actually divided by x + 3. Sample answer: Begin by multiplying two binomials ANSWER: such as (x + 2)(x + 3) which simplifies to x2 + 5x + Sample answer: Sharon; Jamal actually divided by x 6. In order to get a remainder of 3 when divided, add 2 + 3. 3 to the trinomial to get x + 5x + 9. When divided, there will be a remainder of 3. The quotient of two CHALLENGE 44. If a polynomial is divided by a binomial and the remainder is 0, what does this tell polynomials is . you about the relationship between the binomial and the polynomial? ANSWER: SOLUTION: If a polynomial divided by a binomial has no Sample answer: remainder, then the polynomial has two factors: the binomial and the quotient. 47. CCSS ARGUMENTS Identify the expression that does not belong with the other three. Explain your ANSWER: reasoning. The binomial is a factor of the polynomial.

45. REASONING Review any of the division problems in this lesson. What is the relationship between the degrees of the dividend, the divisor, and the quotient?

SOLUTION: Sample answer: For example: SOLUTION: does not belong with the other three. The other

three expressions are polynomials. Since the denominator of contains a variable, it is not a polynomial.

ANSWER:

does not belong with the other three. The other three expressions are polynomials. Since the

denominator of contains a variable, it is not a In this exercise, the degree of the dividend is 2. The polynomial. degree of the divisor and the quotient are each 1. The degree of the quotient plus the degree of the 48. WRITING IN MATH Use the information at the divisor equals the degree of the dividend. beginning of the lesson to write assembly instruction using the division of polynomials to make a paper ANSWER: cover for your textbook. Sample answer: The degree of the quotient plus the degree of the divisor equals the degree of the SOLUTION: dividend. Sample answer: By dividing the area of the paper 140x2 + 60x by the height of the book jacket 10x, the 46. OPEN ENDED Write a quotient of two polynomials quotient of 14x + 6 provides the length of the book for which the remainder is 3. jacket. The front and back cover are each 6x units SOLUTION: long and the spine is 2x units long. Then, subtracting 14x, we are left with 6 inches. Half of this length is Sample answer: Begin by multiplying two binomials the width of each flap. such as (x + 2)(x + 3) which simplifies to x2 + 5x + 6. In order to get a remainder of 3 when divided, add ANSWER: 2 2 3 to the trinomial to get x + 5x + 9. When divided, Sample answer: By dividing 140x + 60x by 10x, the there will be a remainder of 3. The quotient of two quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 polynomials is . inches. Half of this length is the width of each flap.

ANSWER: 49. An office employs x women and 3 men. What is the ratio of the total number of employees to the number Sample answer: of women?

A 47. CCSS ARGUMENTS Identify the expression that

does not belong with the other three. Explain your B reasoning.

SOLUTION: SOLUTION: Total number of employees: x + 3 Number of women: x does not belong with the other three. The other Ratio of the total number of employees to the number three expressions are polynomials. Since the of women is:

denominator of contains a variable, it is not a polynomial. The correct choice is A.

ANSWER: ANSWER: A does not belong with the other three. The other 50. SAT/ACT Which polynomial has degree 3? three expressions are polynomials. Since the A x3 + x2 –2x4 2 denominator of contains a variable, it is not a B –2x – 3x + 4 C polynomial. 3x – 3 D x2 + x + 123 3 48. WRITING IN MATH Use the information at the E 1 + x + x beginning of the lesson to write assembly instruction using the division of polynomials to make a paper SOLUTION: cover for your textbook. To determine the degree of a polynomial, look at the

SOLUTION: term with the greatest exponent.

Sample answer: By dividing the area of the paper The correct choice is E. 2 140x + 60x by the height of the book jacket 10x, the ANSWER: quotient of 14x + 6 provides the length of the book E jacket. The front and back cover are each 6x units long and the spine is 2x units long. Then, subtracting 51. GRIDDED RESPONSE In the figure below, 14x, we are left with 6 inches. Half of this length is m + n + p = ? the width of each flap.

ANSWER: 2 Sample answer: By dividing 140x + 60x by 10x, the quotient of 14x + 6 provides the length of the book jacket. Then, subtracting 14x, we are left with 6 inches. Half of this length is the width of each flap. SOLUTION: 49. An office employs x women and 3 men. What is the Sum of the exterior angles of a triangle is . ratio of the total number of employees to the number So, of women? ANSWER: A 360

B 52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = 2 F 2x C H –10x2 + 17x 2 G –10x D J 2x2 + 17x SOLUTION: SOLUTION: Total number of employees: x + 3 Number of women: x Ratio of the total number of employees to the number

of women is:

The correct choice is H.

The correct choice is A. ANSWER: H ANSWER: A Simplify. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) 50. SAT/ACT Which polynomial has degree 3? A x3 + x2 –2x4 SOLUTION: 2 B –2x – 3x + 4 Use the Distributive Property and then combine like terms. C 3x – 3

2 3 D x + x + 12 3 E 1 + x + x SOLUTION: To determine the degree of a polynomial, look at the term with the greatest exponent. The correct choice is E. ANSWER: 3x3 + 2x2 + x + 4 ANSWER: E 3 2 54. (2y – 3y + 8) + (3y – 6y) 51. GRIDDED RESPONSE In the figure below, SOLUTION: m + n + p = ?

ANSWER: 2y 3 + 3y 2 – 9y + 8

SOLUTION: 55. 4a(2a – 3) + 3a(5a – 4) Sum of the exterior angles of a triangle is . SOLUTION: So, Use the Distributive Property and then combine like terms. ANSWER: 360

52. (–4x2 + 2x + 3) – 3(2x2 – 5x + 1) = ANSWER: 2 2 F 2x 23a – 24a H –10x2 + 17x 2 56. (c + d)(c – d)(2c – 3d) G –10x SOLUTION: J 2x2 + 17x Use the FOIL method and then combine like terms. SOLUTION:

ANSWER: 2c3 – 3c2d – 2cd 2 + 3d 3 The correct choice is H. 2 2 3 57. (xy) (2xy z) ANSWER: H SOLUTION: Apply the properties of exponents to simplify the Simplify. expression. 3 2 3 53. (5x + 2x – 3x + 4) – (2x – 4x) SOLUTION: Use the Distributive Property and then combine like ANSWER: terms. 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER: 3x3 + 2x2 + x + 4

3 2 54. (2y – 3y + 8) + (3y – 6y) ANSWER: SOLUTION:

ANSWER: 59. LANDSCAPING Amado wants to plant a garden 3 2 and surround it with decorative stones. He has 2y + 3y – 9y + 8 enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to 55. 4a(2a – 3) + 3a(5a – 4) cover no more than 240 square feet. What could the SOLUTION: width of his garden be? Use the Distributive Property and then combine like SOLUTION: terms. Let x be the length and y be the width of the garden.

ANSWER: 2 23a – 24a

56. (c + d)(c – d)(2c – 3d) SOLUTION: Use the FOIL method and then combine like terms. The solution of the inequality is

. Since y represent the width of the garden and the sum of the length and ANSWER: width is 34, the width of the garden is

2c3 – 3c2d – 2cd 2 + 3d 3 ANSWER: 2 2 3 57. (xy) (2xy z) 0 to 10 ft or 24 to 34 ft SOLUTION: Solve each equation by completing the square. Apply the properties of exponents to simplify the 2 expression. 60. x + 6x + 2 = 0 SOLUTION:

ANSWER: 8x5y 8z 3

2 –2 2 2 58. (3ab ) (2a b) SOLUTION:

ANSWER:

–3 ±

ANSWER: 2 61. x – 8x – 3 = 0

SOLUTION:

59. LANDSCAPING Amado wants to plant a garden and surround it with decorative stones. He has enough stones to enclose a rectangular garden with a perimeter of 68 feet, but he wants the garden to cover no more than 240 square feet. What could the width of his garden be? SOLUTION: ANSWER: Let x be the length and y be the width of the garden.

62. 2x2 + 6x + 5 = 0 SOLUTION:

The solution of the inequality is . Since y represent the width of the garden and the sum of the length and width is 34, the width of the garden is

ANSWER: 0 to 10 ft or 24 to 34 ft ANSWER: Solve each equation by completing the square. 60. x2 + 6x + 2 = 0 SOLUTION: State the consecutive integers between which the zeros of each quadratic function are located.

63. SOLUTION: The sign of f (x) changes between the x values –6 and –5, and between –3 and –2. So the zeros of the quadratic function lies between –6 and –5, and between –3 and –2.

ANSWER:

between –6 and –5; between –3 and –2 ANSWER:

–3 ±

2 64. 61. x – 8x – 3 = 0 SOLUTION: SOLUTION: The sign of f (x) changes between the x values between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and between 2 and 3.

ANSWER: between 0 and 1; between 2 and 3

ANSWER: 65. SOLUTION: 2 62. 2x + 6x + 5 = 0 The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the SOLUTION: zeros of the quadratic function lies between –1 and 0, and between 2 and 3.

ANSWER: between –1 and 0; between 2 and 3

ANSWER: maximize income. Then find the maximum income. SOLUTION: Let M be the lawns mowed and L be the small landscaping jobs. Since the landscaper will do at most State the consecutive integers between which 3 landscaping jobs per day, L ≤ 15. Since he works the zeros of each quadratic function are located. 10 hours per day, 5 days a week, he can do at most 20 mowing jobs per day or 100 per week. So, M ≤ 100.

63. SOLUTION: If the landscaper does 3 landscaping jobs per day, he The sign of f (x) changes between the x values –6 can only do 11 mowing jobs per day and work a total and –5, and between –3 and –2. So the zeros of the of 10 hours per day. So, for the week, if he does 15 quadratic function lies between –6 and –5, and landscaping jobs, he can mow 55 lawns. between –3 and –2. To maximize the earnings, write a function that ANSWER: relates the number and charge for each type of job. between –6 and –5; between –3 and –2 Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M + 125L. Next, substitute in the key values of M and L to determine the earnings for each combination of 64. jobs.

SOLUTION: (M, L) F(M, L) = 35M + F(M, L) The sign of f (x) changes between the x values 125L between 0 and 1, and between 2 and 3. So the zeros of the quadratic function lies between 0 and 1, and (100, 0) F(M, L) = 35(100) + 3500 between 2 and 3. 125(0) ANSWER: (55, 15) F(M, L) = 35(55) + 125 3800 between 0 and 1; between 2 and 3 (15)

(0, 15) F(M, L) = 35(0) + 125 1875 65. (15) SOLUTION: 15 landscape jobs and 55 lawns; $3800 The sign of f (x) changes between the x values between –1 and 0, and between 2 and 3. So the ANSWER: zeros of the quadratic function lies between –1 and 0, 15 landscape jobs and 55 lawns; $3800 and between 2 and 3. 2 ANSWER: Find each value if f (x) = 4x + 3, g(x) = –x , and 2 between –1 and 0; between 2 and 3 h(x) = –2x – 2x + 4. 67. f (–6) 66. BUSINESS A landscaper can mow a lawn in 30 minutes and perform a small landscape job in 90 SOLUTION: minutes. He works at most 10 hours per day, 5 days Substitute –6 for x in f (x). per week. He earns $35 per lawn and $125 per landscape job. He cannot do more than 3 landscape jobs per day. Find the combination of lawns mowed and completed landscape jobs per week that will maximize income. Then find the maximum income. SOLUTION: ANSWER: Let M be the lawns mowed and L be the small –21 landscaping jobs. Since the landscaper will do at most 3 landscaping jobs per day, L ≤ 15. Since he works 68. g(–8) 10 hours per day, 5 days a week, he can do at most SOLUTION: 20 mowing jobs per day or 100 per week. So, M ≤ 100. Substitute –8 for x in g(x).

If the landscaper does 3 landscaping jobs per day, he can only do 11 mowing jobs per day and work a total of 10 hours per day. So, for the week, if he does 15 landscaping jobs, he can mow 55 lawns. ANSWER: –64 To maximize the earnings, write a function that relates the number and charge for each type of job. 69. h(3) Since he charges $35 for each lawn mowed and $125 for each small landscaping job, F(M, L) = 35M SOLUTION: + 125L. Next, substitute in the key values of M and Substitute 3 for x in h(x). L to determine the earnings for each combination of jobs.

(M, L) F(M, L) = 35M + F(M, L) 125L

(100, 0) F(M, L) = 35(100) + 3500 ANSWER: 125(0) –20 (55, 15) F(M, L) = 35(55) + 125 3800 70. f (c) (15) SOLUTION: (0, 15) F(M, L) = 35(0) + 125 1875 Substitute c for x in f (x). (15)

15 landscape jobs and 55 lawns; $3800

5-2 DividingANSWER: Polynomials ANSWER: 15 landscape jobs and 55 lawns; $3800 4c + 3

Find each value if f (x) = 4x + 3, g(x) = –x2, and 71. g(3d) h(x) = –2x2 – 2x + 4. SOLUTION: 67. f (–6) Substitute 3d for x in g(x). SOLUTION:

Substitute –6 for x in f (x).

ANSWER: 2 –9d

ANSWER: 72. h(2b + 1) –21 SOLUTION: Substitute 2b + 1 for x in h(x). 68. g(–8) SOLUTION: Substitute –8 for x in g(x).

ANSWER: ANSWER: –8b2 – 12b –64 69. h(3) SOLUTION: Substitute 3 for x in h(x).

ANSWER: –20 70. f (c) SOLUTION: Substitute c for x in f (x).

ANSWER: 4c + 3 71. g(3d) eSolutionsSOLUTION: Manual - Powered by Cognero Page 17 Substitute 3d for x in g(x).

ANSWER: 2 –9d

72. h(2b + 1) SOLUTION: Substitute 2b + 1 for x in h(x).

ANSWER: –8b2 – 12b