Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 377: 39


The roots of the polynomial function are:\[\left\{ 1,-2,5 \right\}\]

Work Step by Step

Consider the given polynomial $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10$. Determine values of p and q where pare the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial. $p=\pm 1,\pm 2,\pm 5,\pm 10$ $q=\pm 1$ Calculate $\frac{p}{q}$ . $\frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$ By using the Descartes’ rule of sign. The function $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10$ has two sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has two or zero positive real root. Evaluate $f\left( -x \right):$ $\begin{align} & f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10 \\ & f\left( -x \right)={{\left( -x \right)}^{3}}-4{{\left( -x \right)}^{2}}-7\left( -x \right)+10 \\ & =-{{x}^{3}}-4{{x}^{2}}+7x+10 \end{align}$ The function has only one sign change, thus the function has one negative real zero. Test $x=1$ as a root of the polynomial: $\begin{align} & f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10 \\ & f\left( 1 \right)={{\left( 1 \right)}^{3}}-4{{\left( 1 \right)}^{2}}-7\left( 1 \right)+10 \\ & =1-4-7+10 \\ & =0 \end{align}$ Therefore, the factor of the polynomial is $\left( x-1 \right)$. Divide the equation $f\left( x \right)$ by $\left( x-1 \right)$: $\frac{{{x}^{3}}-4{{x}^{2}}-7x+10}{x-1}={{x}^{2}}-3x-10$ Thus, the polynomial can be expressed as $f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)$. Equate $f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)$ to zero. $\begin{align} & \left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)=0 \\ & \left( x-1 \right)\left( x+2 \right)\left( x-5 \right)=0 \\ & x=1,-2,5 \end{align}$ The solution of the set is $\left\{ 1,-2,5 \right\}$.
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