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: 40


The roots of the polynomial $f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10$ are $\left\{ -1,-10 \right\}$.

Work Step by Step

Determine values of p and q where p are 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 $\frac{p}{q}$. $\frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$ According to Descartes’ s rule of signs, the function $f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10$ has zero sign changes. Since the function $f\left( x \right)$ has no variations in sign, the function $f\left( x \right)$ has zero positive real roots. Further, evaluate $f\left( -x \right):$ $\begin{align} & f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10 \\ & f\left( -x \right)={{\left( -x \right)}^{3}}+12{{\left( -x \right)}^{2}}+21\left( -x \right)+10 \\ & =-{{x}^{3}}+12{{x}^{2}}-21x+10 \end{align}$ There are three variations in sign. Thus, there are 3 negative real zeros or $3-2=1$ negative real zero. Test $x=-1$ as a root of the polynomial: $\begin{align} & f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10 \\ & f\left( -1 \right)={{\left( -1 \right)}^{3}}+12{{\left( -1 \right)}^{2}}+21\left( -1 \right)+10 \\ & =-1+12-21+10 \\ & =0 \end{align}$ Therefore $\left( x+1 \right)$ is a factor of the polynomial. Divide the equation $f\left( x \right)$ by $\left( x+1 \right)$: $\frac{{{x}^{3}}+12{{x}^{2}}+21x+10}{x+1}={{x}^{2}}+11x+10$ Thus, the polynomial can be expressed as $f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}+11x+10 \right)$. Equatethe polynomial equal to zero. $\begin{align} & \left( x+1 \right)\left( {{x}^{2}}+11x+10 \right)=0 \\ & \left( x+1 \right)\left( x+1 \right)\left( x+10 \right)=0 \\ & x=-1,-1,-10 \end{align}$ The solution of the polynomial is $\left\{ -1,-10 \right\}$.
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