+0

# help algebra

0
30
1

Given that x=2 is a root of p(x)=x^4-3x^3-18x^2+90x-100, find the other roots (real and nonreal).

Jan 9, 2022

#1
+26
0

By synthetic division, we get that $$p(x) = (x - 2)(x^3 - x^2 - 20x + 50)$$. By Rational Root Theorem, the possible roots of $$x^3 - x^2 - 20x + 50$$ are $$\pm 1, \pm 2, \pm 5, \pm 10, \pm 25$$, and $$\pm 50$$. Testing values, we get that $$x = -5$$ works, so by synthetic division, we get that $$p(x) = (x - 2)(x + 5)(x^2 - 6x + 10)$$.

By the quadratic formula, the roots of $$x^2 - 6x + 10$$ are $$\frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \frac{6 \pm 2i}{2} = 3 \pm i$$. Thus, the roots of $$p(x)$$ are $$\boxed{-5, 2, 3 + i, \text{and } 3 - i}$$.

Jan 10, 2022