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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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