## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Mid-Chapter Check Point - Page 381: 21

#### Answer

The roots of the polynomial equation are $x=1$ and $x=-2$.

#### Work Step by Step

Consider the polynomial equation ${{x}^{3}}-3x+2=0$. Use the rational root theorem to observe: ${{a}_{0}}=2\text{ , }{{\text{a}}_{n}}=1$ The factors of ${{a}_{0}}$ are: 1,2 and the factors of ${{a}_{n}}$ are: 1 Possible roots are given by: $\pm \frac{1,2}{1}$. $\frac{1}{1}$ is a root of the expression, so factor out $x-1$. Compute $\frac{{{x}^{3}}-3x+2}{x-1}$ to get the rest of the equation ${{x}^{2}}+x-2$. \begin{align} & \left( x-1 \right)\left( {{x}^{2}}+x-2 \right)=0 \\ & x=1,-2 \\ \end{align} Hence, the roots of the polynomial equation ${{x}^{3}}-3x+2=0$ are $x=1$ and $x=-2$.

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