## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 379: 77

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#### Work Step by Step

Consider the function: $f\left( x \right)={{x}^{5}}-{{x}^{4}}+{{x}^{3}}-{{x}^{2}}+x-8$ The signs of the coefficients of the polynomial $f\left( x \right)$ are: $\begin{matrix} + & - & + & - & + & - \\ \end{matrix}$ Therefore, the number of the changes in signs is 5. This means the number of positive zeros is either 5, 3, or 1. Consider the function: $f\left( -x \right)=-{{x}^{5}}-{{x}^{4}}-{{x}^{3}}-{{x}^{2}}-x-8$ The signs of the coefficients of the polynomial $f\left( -x \right)$ are: $\begin{matrix} - & - & - & - & - & - \\ \end{matrix}$ Therefore, there are no sign changes of $f\left( x \right)$ and this means that there are no negative roots. Therefore, the function has one positive zero and no negative zeros, which verifies the possible roots obtained using Descartes’ Rule of Signs.

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