# Math 140 P2 Exam #2 Solutions

1. Factor each polynomial completely .

2. Perform the indicated operations . Leave your answer in factored form.

3. Solve the equation 2x^{2} − 8x + 4 = 0 by
completing the square.

2x^{2} − 8x + 4 = 0 2x^{2} −
8x = −4
x^{2} − 4x = −2.

So, m = −4, , and.
Hence, we have

x^{2} − 4x + 4 = −2 + 4 (x − 2)^{2}
= 2. By the square root method

The solution set is.

4. Solve each of the inequalities , and graph their solution set.

Interval notation:

Interval notation: .

5.Find all real solutions to the following equations.

(a) |3x − 5| = 17.

Either 3x − 5 = 17 or 3x − 5 = −17
3x = 22 or 3x = −12

or x = −4. Solution set:.

(b) x(3x − 7) = −1.

x(3x − 7) = −1 3x^{2}
− 7x = −1 3x^{2}
− 7x + 1 = 0.

Using the quadratic formula .

Solution set:
.

By the zero - product property , either x = 2, x = −2, or x = 4.

Solution set: {−2, 2, 4}.

(d) |x^{2} − 7x + 6|= 6.

Either x^{2} − 7x + 6 = 6 or x^{2} − 7x + 6 = −6.

x^{2} − 7x = 0
or x^{2} − 7x + 12 = 0

x(x − 7) = 0 or (x −
4)(x − 3) = 0.

By the zero- product property , either x = 0, x = 7, x = 3, or x = 4.

Solution set: {0, 3, 4, 7}.

6. Express the following complex numbers in the standard form a + bi.

7. Solve the following equations over the complex numbers .

(a) 5x^{2} + 2 = −3x.

5x^{2} + 2 = −3x
5x^{2} + 3x +
2 = 0. Using the quadratic formula,

Solution set:.

By the zero- product property , either x = −2, x = 1, or x^{2} −2x+4 = 0.
Use the quadratic formula

on x^{2} − 2x + 4 = 0 and get

Putting everything together, we have x = −2, x = 1, or
. So the solution set is

8. **Extra Credit** Solve x^{6} − 1 = 0 over the
complex number system. (Suggestion: Factor using the difference

of squares to get started.)

x^{6} − 1 = 0
(x^{3} − 1)(x^{3} + 1) = 0
(x − 1)(x^{2}
+ x + 1)(x + 1)(x^{2} − x + 1) = 0. Using the quadratic

formula on x ^{2} + x + 1 = 0, we get

Using the quadratic formula on x ^{2} − x + 1 = 0,
we get

Solution set:.

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