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: 15

Answer

The zeros of $f\left( x \right)={{x}^{4}}-5{{x}^{2}}+4$ are $1,-1,2,-2$.

Work Step by Step

Let’s first equate $f\left( x \right)$ to $0$ as below. Then, $\begin{align} & {{x}^{4}}-5{{x}^{2}}+4=0 \\ & {{x}^{4}}-4{{x}^{2}}-{{x}^{2}}+4=0 \\ & {{x}^{2}}\left( {{x}^{2}}-4 \right)-\left( {{x}^{2}}-4 \right)=0 \\ & \left( {{x}^{2}}-4 \right)\left( {{x}^{2}}-1 \right)=0 \end{align}$ Now, by using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ , the polynomial can be written as: $\left( x-1 \right)\left( x+1 \right)\left( x-2 \right)\left( x+2 \right)=0$ Put each factor equal to $0$. So, $\left( x-1 \right)=0$ Or, $\left( x+1 \right)=0$ Or, $\left( x-2 \right)=0$ Or, $\left( x+2 \right)=0$ The above equations provide the values of x as: $x=1,-1,2,-2$
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