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: 27


The required ${{n}^{th}}$ degree polynomial function is ${{x}^{3}}-3{{x}^{2}}-15x+125$

Work Step by Step

The general linear equation is $f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)$ , the roots are ${{c}_{1}}=4+3i$ , ${{c}_{2}}=4-3i$ and ${{c}_{3}}=-5$. $\begin{align} & f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\ & ={{a}_{n}}\left( x-4-3i \right)\left( x-4+3i \right)\left( x+5 \right) \\ & ={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \end{align}$ Calculate the value of ${{a}_{n}}$. Use $f\left( 2 \right)=91$ $\begin{align} & f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\ & 91={{a}_{n}}\left( {{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}-15\left( 2 \right)+125 \right) \\ & ={{a}_{n}}\left( 8-12-30+125 \right) \\ & {{a}_{n}}=\frac{91}{91} \\ & =1 \end{align}$ Substitute the value of ${{a}_{n}}$ in the linear equation. $\begin{align} & f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\ & =1\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\ & ={{x}^{3}}-3{{x}^{2}}-15x+125 \end{align}$ Therefore, the polynomial equation is ${{x}^{3}}-3{{x}^{2}}-15x+125$.
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