Factor Each Polynomial. 1. X − 9 SOLUTION: 2. 4A

Home , Square (algebra)

8-8 Differences of Squares

Factor each polynomial. 2 1. x − 9 SOLUTION:

2. 4a2 − 25 SOLUTION:

2 3. 9m − 144 SOLUTION:

4. 2p 3 − 162p SOLUTION:

4 5. u − 81 SOLUTION:

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6. 2d4 − 32f 4 SOLUTION:

4 4 7. 20r − 45n SOLUTION:

8. 256n4 − c4 SOLUTION:

3 2 9. 2c + 3c − 2c − 3 SOLUTION:

10. f 3 − 4f 2 − 9f + 36 SOLUTION:

3 2 11. 3t + 2t − 48t − 32 SOLUTION:

12. w3 − 3w2 − 9w + 27 SOLUTION:

EXTENDED RESPONSE After an accident, skid marks may result from sudden breaking. The formula s2 = d approximates a vehicle’s speed s in miles per hour given the length d in feet of the skid marks on dry concrete. 13. If skid marks on dry concrete are 54 feet long, how fast was the car traveling when the brakes were applied? SOLUTION:

Since the speed cannot be negative, the car was traveling 36 mph when the brakes were applied.

14. If the skid marks on dry concrete are 150 feet long, how fast was the car traveling when the brakes were applied? SOLUTION:

Since the speed cannot be negative, the car was traveling 60 mph when the brakes were applied.

Factor each polynomial. 2 15. q − 121 SOLUTION:

16. r4 − k4 SOLUTION:

4 17. 6n − 6 SOLUTION:

18. w4 − 625 SOLUTION:

2 2 19. r − 9t SOLUTION:

20. 2c2 − 32d2 SOLUTION:

3 21. h − 100h SOLUTION:

22. h4 − 256 SOLUTION:

3 2 23. 2x − x − 162x + 81 SOLUTION:

24. x2 − 4y2 SOLUTION:

4 4 25. 7h − 7p SOLUTION:

26. 3c3 + 2c2 − 147c − 98 SOLUTION:

2 4 4 27. 6k h − 54k SOLUTION:

28. 5a3 − 20a SOLUTION:

3 2 29. f + 2f − 64f − 128 SOLUTION:

30. 3r3 − 192r SOLUTION:

3 31. 10q − 1210q SOLUTION:

32. 3xn4 − 27x3 SOLUTION:

3 5 3 33. p r − p r SOLUTION:

34. 8c3 − 8c SOLUTION:

3 2 35. r − 5r − 100r + 500 SOLUTION:

36. 3t3 − 7t2 − 3t + 7 SOLUTION:

2 37. a − 49 SOLUTION:

38. 4m3 + 9m2 − 36m − 81 SOLUTION:

4 39. 3m + 243 SOLUTION:

40. 3x3 + x2 − 75x − 25 SOLUTION:

3 2 41. 12a + 2a − 192a − 32 SOLUTION:

42. x4 + 6x3 − 36x2 − 216x SOLUTION:

3 2 43. 15m + 12m − 375m − 300 SOLUTION:

44. GEOMETRY The drawing shown is a square with a square cut out of it.

a. Write an expression that represents the area of the shaded region.

b. Find the dimensions of a rectangle with the same area as the shaded region in the drawing. Assume that the dimensions of the rectangle must be represented by binomials with integral coefficients. SOLUTION: a. Use the formula for the area of square A = s · s to write an expression for the area of the shaded region.

b. Write the area of the shaded region in factor form to find two binomial dimensions for a rectangle with the same area.

The dimensions of the rectangle would be (4n + 6) by (4n − 4).

45. DECORATIONS An arch decorated with balloons was used to decorate the gym for the spring dance. The shape of the arch can be modeled by the equation y = −0.5x2 + 4.5x, where x and y are measured in feet and the x-axis represents the floor.

a. Write the expression that represents the height of the arch in factored form.

b. How far apart are the two points where the arch touches the floor?

c. Graph this equation on your calculator. What is the highest point of the arch? SOLUTION: a.

b. Find the two places along the x-axis (the floor) where the height of the arch is 0.

Subtract the two values to find out how far apart they are: 9 – 0 = 9. The two points where the arch touches the floor are 9 ft. apart.

c. Use a graphing calculator to find the height of the arch at its highest point.

[–2, 10] scl: 2 by [–2, 10] scl: 2

The arch is 10.125 ft. high.

46. CCSS SENSE-MAKING Zelda is building a deck in her backyard. The plans for the deck show that it is to be 24 feet by 24 feet. Zelda wants to reduce one dimension by a number of feet and increase the other dimension by the same number of feet. If the area of the reduced deck is 512 square feet, what are the dimensions of the deck? SOLUTION: Let x be the number of feet added and subtracted to each dimension of the deck. Replace A with 512, l with 24 + x, and w with 24 - x.

Since the amount added and subtracted to the dimensions cannot be negative, 8 feet is the amount that is added and subtracted to the dimensions.

24 – 8 = 16 24 + 8 = 32

Therefore, the dimensions of the deck are 16 feet by 32 feet.

2 47. SALES The sales of a particular CD can be modeled by the equation S = −25m + 125m, where S is the number of CDs sold in thousands, and m is the number of months that it is on the market.

a. In what month should the music store expect the CD to stop selling?

b. In what month will CD sales peak?

c. How many copies will the CD sell at its peak? SOLUTION: a. Find the values of m for which S equals 0.

m = 0 or m = 5

The music store should expect the CD to stop selling in month 5.

b. Use a graphing calculator to find the maximum.

[0, 10] scl: 1 by [0, 250] scl: 25

The maximum occurs at the point x = 2.5. So the CD sales should peak in month 2.5.

The maximum on the graph is S = 156.25, where S is measured in the thousands. The peak number of copies sold is 156.25 × 1000 = 156,250 copies.

Solve each equation by factoring. Confirm your answers using a graphing calculator. 48. 36w2 = 121 SOLUTION:

The roots are and or about 1.833 and –1.833.

Confirm the roots using a graphing calculator. Let Y1 = 36w2 and Y2 = 121. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are and .

49. 100 = 25x2 SOLUTION:

The roots are –2 and 2.

Confirm the roots using a graphing calculator. Let Y1 = 100 and Y2 = 25x2. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are –2 and 2.

50. 64x2 − 1 = 0 SOLUTION:

The roots are and or –0.125 and 0.125.

Confirm the roots using a graphing calculator. Let Y1 = 64x2 – 1 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are and .

2 51. 4y − = 0

SOLUTION:

The roots are and or -0.375 and 0.375.

Confirm the roots using a graphing calculator. Let Y1 = 4y 2 – and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are and .

52. b2 = 16

SOLUTION:

The roots are –8 and 8.

Confirm the roots using a graphing calculator. Let Y1 = and Y2 = 16. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are –8 and 8.

2 53. 81 − x = 0

SOLUTION:

The roots are –45 and 45.

Confirm the roots using a graphing calculator. Let Y1 = and Y2 =0. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are –45 and 45.

54. 9d2 − 81 = 0 SOLUTION:

The roots are –3 and 3.

Confirm the roots using a graphing calculator. Let Y1 = 9d2 – 81 and Y2 =0. Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are –3 and 3.

2 55. 4a =

SOLUTION:

The roots are and or –0.1875 and 0.1875.

Confirm the roots using a graphing calculator. Let Y1 = 4a2 and Y2 = . Use the intersect option from the CALC menu to find the points of intersection.

Thus, the solutions are and .

56. MULTIPLE REPRESENTATIONS In this problem, you will investigate perfect square trinomials.

a. TABULAR Copy and complete the table below by factoring each polynomial. Then write the first and last terms of the given polynomials as perfect squares.

b. ANALYTICAL Write the middle term of each polynomial using the square roots of the perfect squares of the first and last terms.

c. ALGEBRAIC Write the pattern for a perfect square trinomial.

d. VERBAL What conditions must be met for a trinomial to be classified as a perfect square trinomial? SOLUTION: a.

In this trinomial, a = 9, b = –24 and c = 16, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 9(16) or 144 with a sum of –24.

Factors of 144 Sum –1, –144 –145 –2, –72 –74 –3, –48 –51 –4, –36 –40 –6, –24 –30 –8, –18 –26 –9, –16 –25 –12, –12 –24 The correct factors are –12 and –12.

In this trinomial, a = 4, b = –20 and c = 25, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 4(25) or 100 with a sum of –20.

Factors of 100 Sum –1, –100 –101 –2, –50 –52 –4, –25 –29 –10, –10 –20

The correct factors are –10 and –10.

In this trinomial, a = 16, b = –20 and c = 9, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 16(9) or 144 with a sum of 24.

Factors of 144 Sum 1, 144 145 2, 72 74 3, 48 51 4, 36 40 6, 24 30 8, 18 26 9, 16 25 12, 12 24

The correct factors are 12 and 12.

In this trinomial, a = 25, b = 20 and c = 4, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 25(4) or 100 with a sum of 20.

Factors of 100 Sum 1, 100 101 2, 50 52 4, 25 29 10, 10 20

The correct factors are 10 and 10.

b. See table.

c. (a + b)(a + b) = a2 + 2ab + b2 and (a − b)(a − b) = a2 − 2ab + b2

d. The first and last terms must be perfect squares and the middle term must be 2 times the square roots of the first and last terms.

57. ERROR ANALYSIS Elizabeth and Lorenzo are factoring an expression. Is either of them correct? Explain your reasoning.

SOLUTION: Lorenzo is correct.

2 2 Elizabeth’s answer multiplies to 16x − 25y . The exponent on x should be 4.

58. CHALLENGE Factor and simplify 9 − (k + 3)2, a difference of squares. SOLUTION:

2 2 Thus 9 − (k + 3) factor to [3 + (k + 3)][3 − (k + 3)] and simplifies to −k − 6k.

59. CCSS PERSEVERANCE Factor x16 − 81. SOLUTION:

60. REASONING Write and factor a binomial that is the difference of two perfect squares and that has a greatest common factor of 5mk. SOLUTION: Begin with the GCF of 5mk and the difference of squares. 2 2 5mk(a − b )

Distribute the GCF to obtain the binomial of 5mka2 − 5mkb2.

2 2 2 2 5mka − 5mkb = 5mk(a − b ) = 5mk(a − b)(a + b)

61. REASONING Determine whether the following statement is true or false. Give an example or counterexample to justify your answer.

All binomials that have a perfect square in each of the two terms can be factored. SOLUTION: false; The two squares cannot be added together in the binomial.

For example, a2 + b2 cannot be factored.

62. OPEN ENDED Write a binomial in which the difference of squares pattern must be repeated to factor it completely. Then factor the binomial. SOLUTION: If you plan to repeat the the factoring twice, you need to find the difference of two terms to the 4th power.

63. WRITING IN MATH Describe why the difference of squares pattern has no middle term with a variable. SOLUTION: When the difference of squares pattern is multiplied together using the FOIL method, the outer and inner terms are opposites of each other. When these terms are added together, the sum is zero.

Consider the example .

2 64. One of the roots of 2x + 13x = 24 is −8. What is the other root?

SOLUTION:

The correct choice is B.

65. Which of the following is the sum of both solutions of the equation x2 + 3x = 54?

F −21

G −3

H 3

J 21 SOLUTION:

–9 + 6 = –3.

The correct choice is G.

2 66. What are the x-intercepts of the graph of y = −3x + 7x + 20?

A , −4

B , −4

C , 4

D , 4

SOLUTION: To find the x-intercepts, find the zeros or roots of the related equation.

The correct choice is C.

67. EXTENDED RESPONSE Two cars leave Cleveland at the same time from different parts of the city and both drive to Cincinnati. The distance in miles of the cars from the center of Cleveland can be represented by the two equations below, where t represents the time in hours.

Car A: 65t + 15 Car B: 60t + 25

a. Which car is faster? Explain.

b. Find an expression that models the distance between the two cars.

c. How far apart are the cars after 2 hours?

SOLUTION: a. The slope represents the speed of the car. Car A has a speed of 65 mph and Car B has a speed of 60 mph. Thus, Car A is traveling at a greater speed.

c. Substitute in for t.

After hours, the cars are 2.5 miles apart.

Factor each trinomial, if possible. If the trinomial cannot be factored using integers, write prime. 68. 5x2 − 17x + 14 SOLUTION: In this trinomial, a = 5, b = –17 and c = 14, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 5(14) or 70, and look for the pair of factors with a sum of –17.

Factors of 70 Sum –2, –35 –37 –5, –14 –19 –7, –10 –17

The correct factors are –7 and –10.

2 69. 5a − 3a + 15 SOLUTION: In this trinomial, a = 5, b = –3 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 5(15) or 75, and look for the pair of factors with a sum of –3.

Factors of 75 Sum –3, –25 –28 –5, –15 –20

There are no factors of 75 with a sum of –3.

So, the trinomial is prime.

70. 10x2 − 20xy + 10y2 SOLUTION: In this trinomial, a = 10, b = –20 and c = 10, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 10(10) or 100, and look for the pair of factors with a sum of –20.

Factors of 100 Sum –2, –50 –52 –5, –20 –25 –10, –10 –20

The correct factors are –10 and –10.

Solve each equation. Check your solutions. 2 71. n − 9n = −18 SOLUTION:

The roots are 3 and 6. Check by substituting 3 and 6 in for n in the original equation.

and

The solutions are 3 and 6.

72. 10 + a2 = −7a SOLUTION:

The roots are –2 and –5. Check by substituting –2 and –5 in for a in the original equation.

and

The solutions are –2 and –5.

2 73. 22x − x = 96 SOLUTION:

The roots are 6 and 16. Check by substituting 6 and 16 in for x in the original equation.

and

The solutions are 6 and 16.

74. SAVINGS Victoria and Trey each want to buy a scooter. In how many weeks will Victoria and Trey have saved the same amount of money, and how much will each of them have saved?

SOLUTION:

Victoria and Trey will have saved the same amount of money at the end of week 3.

25 + 5(3) = 40

The amount of money they will have saved is $40.

Solve each inequality. Graph the solution set on a number line. 75. t + 14 ≥ 18 SOLUTION:

The solution is {t|t ≥ 4}.

76. d + 5 ≤ 7 SOLUTION:

The solution is {d|d ≤ 2}.

77. −5 + k > −1 SOLUTION:

The solution is {k|k > 4}.

78. 5 < 3 + g SOLUTION:

The solution is {g|g > 2}.

79. 2 ≤ −1 + m SOLUTION:

The solution is {m|m ≥ 3}.

80. 2y > −8 + y SOLUTION:

The solution is {y|y > −8}.

81. FITNESS Silvia is beginning an exercise program that calls for 20 minutes of walking each day for the first week. Each week thereafter, she has to increase her daily walking for the week by 7 minutes. In which week will she first walk over an hour a day? SOLUTION: Silvia's time exercising can be modeled by an arithmetic sequence. 20, 27, 34, 41, 48, 55, ...

a1 is the initial value or 20. d is the difference or 7.

Substitute the values into the formula for the nth term of an arithmetic sequence:

The first week in which she will walk over an hour a day is the seventh week.

Find each product. 2 82. (x − 6) SOLUTION:

83. (x − 2)(x − 2) SOLUTION:

84. (x + 3)(x + 3) SOLUTION:

85. (2x − 5)2 SOLUTION:

2 86. (6x − 1) SOLUTION:

87. (4x + 5)(4x + 5) SOLUTION: Factor each polynomial. 2 1. x − 9 SOLUTION:

2. 4a2 − 25 SOLUTION:

2 3. 9m − 144 SOLUTION:

4. 2p 3 − 162p SOLUTION:

8-8 Differences of Squares

4 5. u − 81 SOLUTION:

6. 2d4 − 32f 4 SOLUTION:

4 4 7. 20r − 45n SOLUTION:

8. 256n4 − c4 SOLUTION:

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