1. Simplify completely. The final answer should have positive exponent only.

2 2 a. x 2 yz 3   2x 2 y 3 z 4  =

When an exponent is on the outside of 4 2 6 2 4 6 8 x y z  2 x y z  = the parenthesis, you multiply exponents.

 x 4 y 2  x 4  Move the negative exponents to make    6 2 = them positive. They move from top to  z   2 y 6 z 8       bottom or bottom to top. x 4 y 2 x 4 Since the fractions are multiplying, I can = 4z 6 y 6 z 8 put them together. I also squared the -2 to get 4.

8 x When variables are both on top or both 4 14 4y z on bottom, you add exponents. When one is on top and the other is on bottom, subtract the exponents. The final answer then goes wherever the higher exponent was to begin with. 2  2a 2b 4  b.   =  1 3 2   3 a b 

22 a 4b8 When an exponent is on the outside of = 32 a 6b 4 the parenthesis, you multiply exponents. Move the negative exponents to make b8 2 2 4 6 4 = them positive. They move from top to 2 3 a a b bottom or bottom to top.

b 4 When variables are both on top or both 36a10 on bottom, you add exponents. When one is on top and the other is on bottom, subtract the exponents. The final answer then goes wherever the higher exponent was to begin with. I also multiplied the numbers in the denominator

2 4 2     2a 3 b 3  c.  1 3 2  =     3   3 2 a 2 b 

4 8  22 a 3 b 3 When an exponent is on the outside of = 4 the parenthesis, you multiply exponents. 3a 3b 3

8 Move the negative exponents to make b 3 them positive. They move from top to 4 4 = 22 3a 3 a 3b 3 bottom or bottom to top.

4 When variables are both on top or both b 3 on bottom, you add exponents. When 13 one is on top and the other is on bottom, 12a 3 subtract the exponents. The final answer then goes wherever the higher exponent was to begin with. 2. Compute the following expressions by hand and write the final answer in scientific notation. You may check your answer on the calculator.

6107 3109  a. = 5106 

18102 = I can multiply the top. The 6 and 3 5106 multiply to give 18. When you multiply the tens, you add the exponents.

3.6104 Now I can divide the 18 and 5 to get 3.6. I subtract the exponents with the tens to get 2 – 6 = -4. Since there is only one number on the left of the decimal, I can leave it. b. 9.1103 8.2105  = I multiply the 9.1 and 8.2 to get 74.62. I 2 add the exponents to get 2 74.62 10 = There is more than one number to the

3 left of the decimal so move the decimal 7.46210 once to the left which adds a value to the exponent. If I move it to the right, I subtract a value from the exponent. 3. Simplify completely and write the answer in radical form

3 12m7 n 4 = My first step is to see if any perfect cubes go into 12. If so, I would cube root that value and take it out. None do, so I am done with that step. Next, I divide the 3 (from the cube root) into the

2 3 exponents of the variables. I leave the m n 12mn remainder under the radical. So, 3 goes into 7 twice (so I take out m 2 ) with one remainder (so I leave an m under the radical.) Likewise with the n which gets an absolute value because it is an even root and an odd exponent. 4. Write the radical expressions using rational exponents. Do not simplify. a. 6 8n 4 = A root becomes a denominator in an exponent. So, 8 is to the first power, 1 then the 6 makes the power . The n is 1 2 6 8 6 n 3 4 2 to the 4th, so it becomes which is . 6 3 1 b. = 6

1 Since the root doesn’t have a number we 1 know it is a square root. So the root 6 2 becomes an exponent to the half.

1  Since the value was on the bottom of a 6 2 fraction, it becomes a negative exponent when we move it out from the bottom.

5. Write the expressions using radicals. Do not simplify.

3 5 3 a. 15 5 = 15

3  3  4x  4  3  b.    4   The top value of an exponent stays,  3   4x  while the bottom becomes the root. The negative exponent makes the 3 2   4 3 fraction flip. The 4 on the bottom d.  3  = 4 3 2  w  w becomes the root while the 3 stays with   the fraction as the exponent. The 2/3 makes it a third root with an exponent of 2. Then, the whole thing is to the third power with a fourth root around it.

6. Simplify completely and write the answer in radical form Rationalize the denominator.

54a 5b 4 15 n 2  3n 2 = a. = 25c 2 12 n 2 9 6a 5b 4 = 25c 2 First I will break up the problem so both the top and bottom get the square root.

To reduce the 54, I find the biggest 3 6a 5b 4 = perfect square (9) that goes into it and 5 c 2 break it down as a multiplication problem.

The square root of 9 is 3, so I take out a 3. Likewise on the bottom, the square root of 25 is 5, so I take that out of the 3a 2b 2 6a root. 5c Since it is a square root, I divide the exponents by 2. When I divide, the whole numbers come out with the variable and the remainder stays under the radical. Any time you take an even 2 2 b.  3 50n  18n = root and the answer for a variable is an odd variable, you put absolute values  3 25 2n 2  9 2n 2 = around it when it comes out of the root.

 3 5 n 2  3n 2 = To reduce the 50 and the 18, I find the biggest perfect square that goes into each one. 50 breaks up into 25 x 2 and 18 into 9 x 2.

Now I can take the square root of the 25 (which is 5) and pull it out. Likewise, I pull out the square root of 9 (which is 3). Since it is a square root, I divide the I can now multiply the outside (-3 x 5) exponents by 2. Since it is an even root and then combine like terms. The and an odd exponent answer, they get radical does not change because it is the absolute value bars. like term. 7. Multiply (3  2 x) by its conjugate and simplify.

Conjugate: 3  2 x  First, to find the conjugate, I rewrite the expression and change the middle sign.

Multiply: 3  2 x 3  2 x  Next, I multiply them. To do this, I foil it out. I get -4x at the end because -2 x 2 9  6 x  6 x  4x is -4 and x times itself is just x.

9  4x I now combine like terms. The middle cancels leaving me with 9 – 4x.

8. Simplify completely and write the answer in radical form.

2 x  54 x  7 These are being multiplied, so I distribute the parenthesis (aka FOIL).

Remember that a square root times itself 8x 14 x  20 x  35 gets rid of the root ( x  x  x )

8x  6 x  35 After I foil, I can combine like terms.

9. Rationalize the denominator of the expression.

2  3 a. x  2 First multiply by the conjugate of the 2  3 x  2  denominator which is x  2 . This must x  2 x  2 be done to the top and the bottom.

2 x  4  3x  2 3 FOIL the top and the bottom and x  4 combine like terms. 10 To get rid of a square root in the b. denominator, you need to multiply it by 3 9 itself. In other words, you need two of them. For a cube root, the process is 10 3 9 3 9 similar, but you need three of them. So   3 9 3 9 3 9 multiply the top and bottom by two of the roots.

103 81 This gets rid of the root in the denominator.

9 I multiply the 9’s in the numerator.

Now I need to reduce the radical by finding 103 27 3 the biggest perfect cube that goes into 81. 9 27 is 3 cubed, so I break 81 up into 27 * 3

The cubed root of 27 is 3 so I pull that out to 10 33 3 be multiplied by the 10. 9

103 3 The 3 on top reduces with the 9 on bottom to be 1 and 3. 3

2 5 To get rid of a square root in the c. denominator, you need to multiply it by 6 itself. In other words, you need two of them.

2 5 6 Multiply the top and bottom by the square  root of 6. 6 6 Multiply the radicals on top and try to 2 30 reduce the radical. 30 will not reduce.

6

The 2 and 6 reduce to 1 and 3. 30

3

10. Simplify completely. a. 2a  33a 2  5a 1 6a 3 19a 2 17a  3

6a 3 10a 2  2a  9a 2 15a  3 First distribute the 2a and the -3 through the second parenthesis by multiplying the 2a and -3 by each piece. Combine like terms. b. x 3  3x 2 y  4xy 2  y 3  7x 3  yx 2  9xy 2  y 3  First distribute the negative through the parenthesis by x3  3x 2 y  4xy 2  y 3  7x 3  yx 2  9xy 2  y 3 multiplying everything in the second parenthesis by -1.

 6x3  4x 2 y 13xy 2 Combine like terms. c. 4x  6y2 When something is squared it is multiplied by itself. So I write the 4x – 6y out twice to multiply it. 4x  6y4x  6y

16x 2  24xy  24xy  36y 2 I FOIL the parenthesis.

Combine like terms. 16x 2  48xy  36y 2

11. Factor completely. a. 6x 4 11x 3 10x 2 First find your GCF (greatest common factor) and factor it out. x 2 6x 2 11x 10 Now factor what is left in the parenthesis x 2 3x  22x  5 using the box method or trial and error. b. 27y 4  8y First find your GCF (greatest common factor) and factor it out.

3 y27y  8 Now we have cubes so use the cube root rule: y3y  29y 2  6y  4 x3  y 3  x  yx 2  xy  y 2  c. 3x 2 10x  8

3x  2x  4 Factor using box or trial and error method. d. 2x3 14x 2  20x First find your GCF (greatest common factor) and factor it out. 2xx 2  7x 10 Factor using what multiplies to get 10 and 2xx  5x  2 adds to get -7. e. 10a 2  3a 18

5a  62a  3 Factor using box or trial and error method. f. 16y 4  81

2 2 4y  94y  9 Factor using the difference of squares.

2y  32y  34y 2  9 Factor the first parenthesis using the difference of squares. The second one won’t factor since it is the sum of squares. g. x3  2x 2  9x 18 Since there are four pieces, I will factor by grouping. x 2 x  2 9x  2 Factor the GCF out of the front two terms. Factor the GCF out of the back two terms.

x 2  9x  2 What the front and back shares gets its own parenthesis. What they don’t share gets its own as well. x  3x  3x  2 Factor the difference of squares in the front.

12. Simplify completely.

2x 2  5x 12 x 2 16 a. 2  2 4x  8x  3 2x  7x  3 Since this is a division problem, I need to flip the second fraction and change it to a 2x 2  5x 12 2x 2  7x  3 multiplication problem.  4x 2  8x  3 x 2 16 Factor each set of terms using difference of 2x  3x  4 2x 1x  3 squares and box method or trial and error.  2x  32x 1 x  4x  4 The (2x + 3), (2x + 1), and (x – 4) all cancel since they are on the top and bottom. Since x  3 1 3 they were on the bottom, they can not be = and x  4, , x  4 2 2 equal to zero, so I have to find the restrictions. To do this, set them all equal to zero and solve. 6x 2  x 1 15 b.  2 Factor each set of terms using difference of 6x  3 9x 1 squares, GCF, and box method or trial and error. 2x 13x 1 15  32x 1 3x 13x 1 The 3 and 15 reduce to 1 and 5. The (3x -1) and (2x + 1) cancel since they are on top and 5 1 1 bottom. Since they were on the bottom, they and x  , cannot equal zero. Find the restrictions by 3x 1 3 2 setting them equal to zero and solving.

13. Simplify completely.

4 Rather than looking at this as one big 1 x 2 problem, I am going to separate it into three a. separate problems. I will focus first on 1 6 1  subtracting the top. x x 2 To subtract fractions, I need a common 1 4 denominator. The first fraction has a 1 so Top:  1 x 2 the x 2 in the second fraction is my common denominator. 1x 2  4 x 2  4 = 2 x 2 x 2 The first fraction is missing the x so I 2 multiply the 1 by x and combine the numerators. 1 1 6 Bottom:   1 x x 2 Now to work on the bottom fraction, I need 2 the common denominator of x . 1x2 1x 6 x 2  x  6 = 2 2 To combine the fractions, the first one needs x x 2 x , the second one needs an x, and the third x 2  4 x 2  x  6 one needs nothing. Combine: 2  2 x x Now I will combine the two parts by dividing. When you divide fractions you x 2  4 x 2 multiply by the reciprocal of the second  x 2 x 2  x  6 fraction.

2 Factor each piece of the fractions. The (x + x  2x  2 x 2 x 2  2) and cancel top to bottom. You can x x  3x  2 not have a zero in the bottom, so I have to x 2 x  2 define restrictions by setting x + 2 and and x  0,2 equal to zero and finding my restrictions x  3 5 2 60 b.   x  2 x 2  2x  4 x 3  8 5 2  60 To add or subtract fractions I need a   common denominator. To do this I must x  2 x 2  2x  4 x  2 x 2  2x  4    first factor my denominators. I factored the last one using the cube root rule (see 11b for 5x 2  2x  4 2x  2 60 help.)

x  2x 2  2x  4 My common denominator is 2 2 x  2x  2x  4. The first fraction is 5x 10x  20  2x  4  60 2 2 missing x  2x  4 . The second one is x  2x  2x  4 missing x + 2. Multiply each numerator by what it is missing. 5x 2  8x  36

x  2x 2  2x  4

Combine like terms. Since the numerator can’t factor, you are finished. 14. Solve the following equations. To solve an equation with fractions, get a common denominator for the equation. 2 m 2 a.   Since the first two fractions had a 3 and the 3 3 m last one had an m, the common denominator is 3m. Multiply each numerator by what 2m  mm 23 was missing from the common denominator. 3m

2m  m2  6 Once you have a common denominator, you can get rid of it to solve the numerator. This

2 is a quadratic, so get everything on one side m  2m  6  0 to solve for the zero’s.

2   22  416 This one won’t factor, so use the quadratic equation. 21

2   20

2 Reduce the radical by finding the biggest perfect square that goes into 20 which 2   4 5 2  2i 5  makes it 4 * 5. 2 2 The square root of 4 is 2 and pull an i out since there was a negative. Reduce the m  1 i 5 fraction by dividing all parts by 2. 2 3 25x  6  33x  8 b. x  6  x  8 3 5 15

10x  6  9x  8 The least common multiple between 3 and 5 10x  60  9x  72 is 15, so that is the denominator. Multiply each numerator by what it was missing. x  12 Once you have a common denominator, you To solve an equation with fractions, get a can get rid of it to solve the numerator. common denominator for the equation. Multiply the front numbers on each part. Distribute the front numbers and solve for x. c.  4a 1 3a  72a  3 Distribute the front numbers (-4 and -7) through the parenthesis.  4a  4  3a  14a  21 Combine like terms on each side of the equal sign.  7a  4  14a  21 Move the -14a to the left by adding 14a and 7a  25 the -4 to the right by adding 4.

25 Divide by 7 to solve for a. a  7 d.  6x  4 66  3x  9x  2 Distribute the numbers (-6 and -9) through  6x  24  66  3x  9x 18 the parenthesis. Combine like terms on each side of the equal sign.  6x  42  6x 18 Move the -6x on the right to the left by adding 6x 42  18 You are left with a statement without x’s x  NS that is not true. So this is not solvable.

1 1 1 e. x  3  2 x  3  2 x  3  2 2 2 2

1 1 x  1 x  5 2 2 Get rid of the absolute value and set up two x  2,10 equations. One is set to the positive of the right side and one is set to the negative of the right side. To solve an absolute value equation, make sure the absolute value portion is by itself on Subtract 3 from both sides. the left of the equation.

Solve by multiplying both sides by 2. 15. Rework the contextual problems in HW 04. One example is shown.

A train travels 150 km at a speed of 40 km per hour. A. Write an equation in terms of t describing the distance the train travels in t time. B. How long will it take the train to travel 300 km? C. If the train travels 240 km in 3 hours how fast was the train traveling in km per hour.

40km 150km Since it is traveling at 40 km per hour, set it A.  1hr (t)hours up as a fraction with 40 km in the numerator and 1 hour in the denominator. This is the left side of the equation.

The right side of the equation lets you set up what you are looking for. You want to keep km in the numerator just like on the left side, so 150 km is there. Likewise the hours stay in the denominator so 7 goes there.

B. 40t = 150 Now cross multiply. So the 40 is multiplied by t and 1 by 150.

t = 7.5 hours Solve by dividing by 40. Once again, set up each portion of the 240km xkm C.  problem as a fraction on one side of the 3hr 1hours equal sign. So 240 km per 3 hr becomes a fraction. X km per one hour is also a fraction on the other side.

240  3x Cross multiply to get 240 times 1 and 3 times x. 80 kph Divide both sides by 3. 16. Solve sr  3  4s for s. 3 s  r  4 sr  4s  3

sr  4  3 To solve for s, we must first get all the s’s on one side. So, subtract 4s from both sides. Factor out the s. Get s by itself by dividing both sides by r-4 Solve and graph the following inequalities and express the answer in interval notation. Be able to express the answer in interval and set notation.

17. 3  3  2x  1 To solve an absolute value problem, get the absolute value by itself. First subtract 3.

3  3  2x  2 Now divide by the negative in the front. This flips the inequality. 3  2x  2 Now that the absolute value is by itself, get rid of the absolute value and set up two 3  2x  2 3  2x  2 inequalities. The first keeps everything the same. The second flips the inequality and 2x  1 2x  5 adds a negative to the other side.

1 5 Solve each inequality by subtracting the 3 x   x   and dividing by 2. 2 2

To graph it, set up a number line with the smaller number on the left. Since it is less than the -1/2, shade to the left of that value. 5 1   Greater than -5/2 shades to the right. Equal 2 2 signs tell you to color in the ends.

 5 1 To write in set notation, put an x bar in the Set : x   x    front. Set up the inequalities with the x in  2 2 the middle.

To write in interval notation, put the smaller  5 1 number, larger number. There are brackets Interval:  ,  2 2 because of the equal signs.

1 2x 7 18.  2   5   x  7 3 2

7  6  1 2x  15  7 2  7  2x  14  7  This is the answer to graph. Set: x   x  7 2   Put the smaller number on the left of the number line and the larger on the right.  7  Shade in between since x is between them. Interval:  ,7  2  Don’t color in either end since it is not equal to on either side.

To write in set notation, put an x bar in the First get rid of the division by multiplying front. Set up the inequalities with the x in all parts by 3. the middle.

To write in interval notation, put the smaller Now get rid of the 1 by subtracting 1 from number, larger number. There are brackets all parts. because of the equal signs. Divide everything by 2.

19. 3x  3  2x  1 7x  9

5x  3  7x  8 First combine like terms on each side.

 2x  5 Now get the 7x to the left by subtracting and the -3 to the right by adding 3. 5 x  Divide by -2 which flips the inequality. 2 To graph the inequality, put the value 5/2 on the line. Since it is less than, shade to the 5 left. The equal sign tells you to color in the circle. 2  5 Set: x x   To write in set notation, put an x bar in the 2   front followed by the inequality.

To write in interval notation, put the smaller  5 number, larger number with the smallest Interval:  , number being negative infinity. There is  2 always a parenthesis with infinity and a bracket on 5/2 because of the equal signs. 20.  5x  5x  6

10x  6  3 Set: x x     5 3 x   5

3  5  3  the right. The equal sign tells you to color Interval:  , in the circle.  5  To write in set notation, put an x bar in the First move the 5x on the right to the left by front followed by the inequality. subtracting it.

Divide both sides by -10. The inequality To write in interval notation, put the smaller flips because you divided by a negative. number, larger number with the smallest number being -3/5 and the largest being To graph the inequality, put the value -3/5 infinity. There is always a parenthesis with on the line. Since it is greater than, shade to infinity and a bracket on 5/2 because of the equal signs.

21. 4x  611 2x  18 First distribute the -6 through the 4x  66 12x  18 parenthesis. Combine like terms on the left. 16x  66  18 Move the -66 by adding 66. 16x  48 Divide by 16. x  3 To graph the inequality, put the value 3 on the line. Since it is greater than, shade to the right. No equal sign tells you to not color in the circle. Set: x x  6 To write in set notation, put an x bar in the front followed by the inequality.

To write in interval notation, put the smaller Interval: 6, number, larger number with the smallest number being 6 and the larger number being infinity. There is always a parenthesis with infinity and a parenthesis on 6 because of no equal sign