Question

Find the roots of the polynomial equation.
x3 - 2x2 + 10x + 136 = 0

A. –3 ± 5i, –4
B. –3 ± i, 4
C. 3 ± 5i, –4
D. 3 ± i, 4

Answer 1

Use the Rational Roots Theorem to find the possibilities for the first root. All the numbers, both positive and negative, that go into both the 136 and the 1 (on the x^3 term). Those possibilities are +/-1, +/-2, +/-4, +/-8, +/-17, +/-34, +/-68, +/-136. Let's start with the smallest positive one, which is +1, and do synthetic division to find out whether it comes out evenly or not. By that I mean that there is no remainder.

Answer 2

Doing synthetic division, you will find that the first root is x = -4, or (x+4). The leftover polynomial is of the second degree, namely x^2-6x+34. That probably needs the quadratic formula to factor it correctly. So let me do that...

Answer 3

When you do that, you get that the other roots are imaginary, namely 3+/-5i. So these are the roots of your polynomial:\[(x+4)(x+[3+5i])(x+[3-5i])\]The choice you want is C from above. Wish you were here to see all of this!