1.
ANSWERS
INTERMEDIATE ALGEBRA
TEST CHAPTER 7
WEB SITE
1 x
3


5
2 y
4

7
2
= the exponents on the x are
2
3
1
-

2
5 6
=
2
5
+
2
=
11
2
4 the exponents on the y are
1
-
7
2
=
8
2
-
7
2

1
2
THE ANSWER TO THIS PROBLEM IS
11 1 x 2 y 2
2
2. 3
( 1 )(

1
)
3 x
(

2
1
)(

1
)
3 y
( 2 )(

3
1
)
= 3

1
3
1 x
6 y

2
3
= You can not have a negative exponent. THE
ANSWER TO THIS PROBLEM IS
1 x 6
1
2
3 3 y 3 x
3


2
4 7
= x
21
28

8
28
= x
29
28
3.
4.
7
6 x
3
5. Must have positive exponents in the answer.
1
1
=
1
4
or .25
64 3
In problems 6 – 9, find the prime factorization of the coefficients.
6. 40

(2)(2)(2)(5)

THE ANSWER IS

(2xy
3
)
3
5 x
2
7. 405

(3)(3)(3)(3)(5)

THE ANSWER IS

9x
3 y
4
5 xz
8. 27

(3)(3)(3)

THE ANSWER IS

(3xy
3
)
3 x
2
9. 16

5 x + 2 9

5 x - 3 4

5 x =
4
4
5 x
+ (2)(3)
4
5 x
- (3)(2)
5 x
=
5 x
+ 6
5 x
- 6
5 x
=
5 x

PAGE 2
10.
11.
Use FOIL to solve this problem.
2 9 + 8 6 - 5 6 - 20 4 = (2)(3) + 3 6 - (20)(2) = 6 + 3 6 - 40 = -34 + 3 6
You must multiply the denominator with a number and/or a letter to make the denominator a perfect square. YOU CAN NEVER HAVE A RADICAL(square root) IN THE
DENOMINATOR.
6
3 x

3 x
3 x
=
6 3
9 x
2 x
=
6 3
3 x x
=
2 denominator to reduce the fraction.) x
3 x
(You can divide 3 into the numerator and the
12.
3
7 x
=
3
7 x


3
3 x x
=
21 x
9 x
2
=
21 x
( You can not reduce this problem because the 21x is
3 x under the radical and the 3x is outside of the radical.)
13.
2
3

2
3

5
=
(
( 2
7 x
7
=
11
7
x =
11
7
3

2 )( 2
3

5 )( 2
3

5 )
3

5 )
= ( Use FOIL to multiply)
4
2 9

5
9

10
3

4
3

10
3

10
3

25
=
( 2 )(
(
3 )
4 )(

9
3 )
3

10

25
=
6

9
12

3

25
10
=
16


9
13
In problems 14 – 16, you must square both sides of the equation before solving the equations.
14. ( 7 x

5 )
2
= 4
2
15. (1 + x

1 )
2
= ( x

2 )
2
7x + 5 = 16
7x + 5 – 5 = 16 – 5
7x = 11
(1 +
1 + 1 x

1 )(1 + x

1 + 1 x

1 ) = x + 2 x

1 + x + 1 = x + 2
3
1 + 1 + x + 2
2 + x + 2 x x


1
1 = x + 2
= x + 2
2 – 2 + x – x + 2
2 x

1 = x – x + 2 – 2 x

1 = 0
2 x

1 0
=
2 2 x

1 = 0
( x

1 )
2
= 0
2
x + 1 = 0
x + 1 – 1 = 0 –1
x = -1

17.
16.
19.
20.
PAGE 3
(x – 3)
2
= ( 3 x

1 )
2
(x – 3)(x – 3) = 3x + 1 x
2
- 3x – 3x + 9 = 3x + 1 x
2
- 6x + 9 = 3x + 1 x
2
- 6x – 3x + 9 – 1 = 3x – 3x + 1 – 1 x
2
- 9x + 8 = 0 (factor the x
2
- 9x + 8)
(x – 1)(x – 8) = 0 ( set each equation to equal zero) x – 1 = 0 ; x – 8 = 0
x = 1 ; 8 (You need to check you answer to see if these numbers work in the problem.) i 16 = 4ị 18. (i 3 )(i 3 ) = ( i
2
)( 9 ) = 3ị to –1). Therefore 3ị
2
= -3
2
(We know that ị
2
is equal
Use FOIL to work problem. (4)(6) + (4)(2ị) + (6)(-3ị) + (-3ị)(2ị) = 24 + 8ị - 18ị - 6ị
2
14 - 10ị - (6)(-1) = 24 -10ị + 6 = 30 - 10ị
= ị = ị This pattern repeats. 4 units to this pattern, therefore 47

4 = 11 r 3. The remainder ị 2
= -1 is what we are interested in. Since the remainder is 3, ị
3
= -ị, therefor the answer to ị
3
= -ị this problem is -ị . ị 4
= 1