 # 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 2x2 − 8x + 4 = 0 by completing the square.
2x2 − 8x + 4 = 0 2x2 − 8x = −4 x2 − 4x = −2.
So, m = −4, , and . Hence, we have
x2 − 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 3x2 − 7x = −1 3x2 − 7x + 1 = 0. Solution set: . By the zero - product property , either x = 2, x = −2, or x = 4.
Solution set: {−2, 2, 4}.

(d) |x2 − 7x + 6|= 6.
Either x2 − 7x + 6 = 6 or x2 − 7x + 6 = −6. x2 − 7x = 0 or x2 − 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) 5x2 + 2 = −3x.
5x2 + 2 = −3x 5x2 + 3x + 2 = 0. Using the quadratic formula, Solution set: . By the zero- product property , either x = −2, x = 1, or x2 −2x+4 = 0. Use the quadratic formula
on x2 − 2x + 4 = 0 and get Putting everything together, we have x = −2, x = 1, or . So the solution set is 8. Extra Credit Solve x6 − 1 = 0 over the complex number system. (Suggestion: Factor using the difference
of squares to get started.)
x6 − 1 = 0 (x3 − 1)(x3 + 1) = 0 (x − 1)(x2 + x + 1)(x + 1)(x2 − 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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