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: 14


The zeros of $f\left( x \right)={{x}^{3}}-{{x}^{2}}-4x+4$ are $1,2,-2$.

Work Step by Step

First put $f\left( x \right)=0$. So, Consider, ${{x}^{3}}-{{x}^{2}}-4x+4$ Factorize it, $\begin{align} & {{x}^{3}}-{{x}^{2}}-4x+4=0 \\ & {{x}^{2}}\left( x-1 \right)-4\left( x-1 \right)=0 \\ & \left( {{x}^{2}}-4 \right)\left( x-1 \right)=0 \end{align}$ By using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ the polynomial ${{x}^{2}}-4$ is written as: ${{x}^{2}}-4=\left( x-2 \right)\left( x+2 \right)$ So, the provided function factorizes as: $\left( x-1 \right)\left( x-2 \right)\left( x+2 \right)=0$ Put each factor equal to $0$. So, $\left( x-1 \right)=0$ Or, $\left( x-2 \right)=0$ Or, $\left( x+2 \right)=0$ The above equations provide the values of x as, $x=1,2,-2$ The graph of the function passes the x-axis at $x=1,2,-2$ since all the roots have a multiplicity of 1. Since it is an odd degree polynomial (degree=3) and the leading coefficient is 1, the graph falls to the left and rises to the right.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.