November 18, 2010
(Total: 100 points + extra credits: 5 points)
You have to show all the work on the space given. Please be neat! If you need more space, you can use extra sheets of paper provided. (Please put your name on all the sheets and label question numbers.) Not showing your work gets 0 point.
Where appropriate, clearly indicate your final answer by circling it. You are not allowed to get help from your textbook, class notes, calculators, other students, or any other form of outside aid. Do not forget to turn off your cell phone. If you have any question, please ask the instructor. Good luck!
1. (Total: 10 points) Factor the following polynomials
(a) (5 points) 36 x
2 − 49 y
2
36 x
2
− 49 y
2
= (6 x − 7 y )(6 x + 7 y )
(b) (5 points) 25 x
2 − 30 x + 9
25 x
2
− 30 x + 9 = (5 x − 3)
2
1

2. (Total: 10 points) Factor each polynomial completely.
(a) (5 points) 16 x
4 − y
4
16 x
4
− y
4
= (4 x
2
+ y
2
)(4 x
2
− y
2
) = (4 x
2
+ y
2
)(2 x + y )(2 x − y )
(b) (5 points) 16 x
3 − 250
16 x
3
− 250 = 2(8 x
3
− 125) = 2(2 x − 5)(4 x
2
+ 10 x + 25)
2

3. (10 points) Combine and simplify the following rational expression
2 x x − 5
+
3 x + 2 x + 4
· x − 5
5
2 x x − 5
+
3 x + 2 x + 4
· x − 5
5
=
2 x ( x + 4) + ( x − 5)(3 x + 2)
( x − 5)( x + 4)
· x − 5
5
=
2 x 2 + 8 x + 3 x 2 − 15 x + 2 x − 10
5( x + 4)
=
5 x 2 − 5 x − 10
5( x + 4)
= x 2 − x − 2 x + 4
3

4. (10 points) Divide x 3 − x 2 − 5 x + 25 by x + 3.
x
3 − x
2 − 5 x + 25 x + 3
= ( x
2 − 4 x + 7)( x + 3) +
4 x + 3
4

5. (10 points) Solve 2 x 2 + 5 x + 3 = 0.
x =
− 5 ±
√
25 − 24
4
=
− 5 ± 1
4
Solutions: x = − 3
2
, x = − 1.
5

6. (Total: 10 points) Evaluate each radical function when x = 2.
(a) (5 points) f ( x ) =
√ x + 2 f (20) =
√
2 + 2 =
√
4 = 2
(b) (5 points) g ( x ) =
√
3
4 x 2 g (2) =
√
3
16 = 2
√
3
2
6

7. (Total: 10 points) Find the domain of each rational function.
(a) (5 points) f ( x ) = x
2
+4 x − 2
Domain = { x | x = 2 }
(b) (5 points) g ( x ) =
1+ x x +3
Domain = { x | x = − 3 }
7

8. (10 points) Solve p
2 x 2 − 4 =
√
3 x − 2
( p
2 x 2 − 4)
2
= (
√
3 x − 2)
2
2 x
2
− 4 = 3 x − 2 x =
2 x
2
3 ±
− 3 x − 2 = 0
√
9 + 16 3 ± 5
=
4 4
There are two possible solutions: x = 2 and x = − 1
2 solutions.
. We need to check if they are
If x = 2, then p
2 · 2 2
√
8
− 4 =
√
√
3 · 2 − 2
−
√
4 =
4 =
√
6 − 2
4
Therefore x = 2 is a solution.
If x = − 1
2
, then r
1
2
− 4 = r
−
3
2
− 2
Therefore x = −
1
2 is NOT a solution.
r
−
7
2
= r
−
7
2
8

9. (Total: 10 points) Perform each operation and write the result in standard form.
(a) (5 points) (3 + 2 i ) − (1 − 7 i )
3 + 2 i − 1 + 7 i = (3 − 1) + (2 + 7) i = 2 + 9 i
(b) (5 points) (3 − 2 i )(2 + 4 i )
(3 − 2 i )(2 + 4 i ) = 6 + 12 i − 4 i − 8 i
2
= 6 + 8 i + 8 = 14 + 8 i
9

10. (Total: 10 points) Simplify the following radical expressions.
(a) (5 points)
3
√
8 +
√
3
2 − 2
3
= (6 − 2)
√
8 +
√
2 + 5
√
√
3
√
3
2 − 2
2 = 6
2 + (1 + 5)
√
√
2 + 5
2 +
√
3
2
√
3
2 − 2
√
√
3
2 = 4
√
2 + 6
2 + 5
√
3
2
√
3
2
(b) (5 points)
√
3
√
3 −
√
2
·
√
3 +
√
3 +
√
3
√
3 −
√
2
√
2
√
2
=
3 +
3 −
√
2
6
= 3 +
√
6
10

11. (Extra problem: 5 points) Write the following quotient of complex numbers
2 + i
3 − 2 i in standard form.
2 + i
3 − 2 i
·
3 + 2 i
3 + 2 i
=
6 + 4 i + 3 i + 2 i 2
9 − 4 i 2
=
4 + 7 i
13
4
=
13
+
7
13 i
–END–
11