Precalculus (6th Edition) Blitzer

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

Chapter 2 - Mid-Chapter Check Point - Page 381: 19

Answer

The zeros of $f\left( x \right)={{x}^{3}}-2{{x}^{2}}+26x$ are $0,1+5i,1-5i$.

Work Step by Step

First put $f\left( x \right)=0$. So, ${{x}^{3}}-2{{x}^{2}}+26x=0$ Then, function $f\left( x \right)={{x}^{3}}-2{{x}^{2}}+26x$ can be factorized as: $x\left( {{x}^{2}}-2x+26 \right)=0$ The function $a{{x}^{2}}+bx+c$ is a quadratic polynomial and the zeros are found by the formula: $\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Now, compare ${{x}^{2}}-2x+26$ with $a{{x}^{2}}+bx+c$. So, the zeros are: $\begin{align} & x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 26 \right)}}{2} \\ & =\frac{2\pm \sqrt{4-104}}{2} \\ & =\frac{2\pm \sqrt{-100}}{2} \\ & =\frac{2\pm 10i}{2} \end{align}$ So, the zeros of ${{x}^{2}}-2x+26$ are $1+5i,1-5i$ Then the zeros of the provided polynomial are: $x=0,1+5i,1-5i$ The function crosses the x-axis at 0 since it has multiplicity 1. Since the function is an odd-degree polynomial and the leading coefficient is 1, the graph falls to the left and rises to the right. At $x=5$ ${{\left( 5 \right)}^{3}}-2{{\left( 5 \right)}^{2}}+26\left( 5 \right)=205$ At $x=-5$ ${{\left( -5 \right)}^{3}}-2{{\left( -5 \right)}^{2}}+26\left( -5 \right)=-305$
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