Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Cumulative Review Exercises - Page 952: 20



Work Step by Step

Consider $f\left( x \right)={{x}^{3}}-6x+4$ Next, consider $\begin{align} & f\left( 2 \right)={{2}^{3}}-6\times 2+4 \\ & =0 \end{align}$ Therefore, using synthetic division, $x-2$ is a factor of the function $f\left( x \right)$ and hence $\begin{align} & f\left( x \right)={{x}^{3}}-6x+4 \\ & ={{x}^{3}}-2{{x}^{2}}+2{{x}^{2}}-4x-2x+4 \\ & ={{x}^{2}}\left( x-2 \right)+2x\left( x-2 \right)-2\left( x-2 \right) \\ & =\left( x-2 \right)\left( {{x}^{2}}+2x-2 \right) \end{align}$ Therefore $\frac{f\left( x \right)}{x-2}={{x}^{2}}+2x-2$
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