Chapter 4.7, Problem 48WE

To determine

To find zero of f.

Answer to Problem 48WE

The zeros of f are $\text{2,-2,}\frac{1}{2}\text{ and }-\frac{1}{2}$

Explanation of Solution

Given information:

  $f\left(s\right)=4s^4-17s^2+4$

Calculation:

The value of f can be calculated as

  $\begin{array}{l}f\left(s\right)=4s^4-17s^2+4\\ 4s^4-17s^2+4=0\left[\text{write }f\left(s\right)=0\right]\\ 4s^4-\left(s^2+16s^2\right)+4=0\left[\text{write 17}{s}^2=s^2+16s^2\right]\\ 4s^4-s^2-16s^2+4=0\\ s^2\left(4s^2-1\right)-4\left(4s^2-1\right)=0\left[\begin{array}{l}\text{the common factor of first and second term is }{s}^2\text{ and}\\ \text{the common factor of third and fourth term is 4}\end{array}\right]\\ \left(4s^2-1\right)\left(s^2-4\right)=0\\ \left[{\left(2s\right)}^2-{\left(1\right)}^2\right]\left[{\left(s\right)}^2-{\left(2\right)}^2\right]=0\left[\text{difference of squares}\right]\\ \left[\left(2s+1\right)\left(2s-1\right)\right]\left[\left(s+2\right)\left(s-2\right)\right]=0\end{array}$

Therefore,

  $\begin{array}{l}\left(s+2\right)=0\text{ or }\left(2s+1\right)=0\text{ or }\left(s-2\right)\text{=0 or }\left(2s-1\right)\text{=0 }\left[\text{use the zero-product property}\right]\\ s=-2\text{ or }2s=-1\text{ or }s=2\text{ or }2s=1\\ s=-2\text{ or }s=-\frac{1}{2}\text{ or }s=2\text{ or }s=\frac{1}{2}\\ \\ \text{check: }f\left(-2\right)=4{\left(-2\right)}^4-17{\left(-2\right)}^2+4=4\left(16\right)-17\left(4\right)+4=64-68+4=0\\ f\left(2\right)=4{\left(2\right)}^4-17{\left(2\right)}^2+4=4\left(16\right)-17\left(4\right)+4=0\\ f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^4-17{\left(-\frac{1}{2}\right)}^2+4=4\left(\frac{1}{16}\right)-17\left(\frac{1}{4}\right)+4=\frac{1}{4}-\frac{17}{4}+\frac{16}{4}=0\\ f\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^4-17{\left(\frac{1}{2}\right)}^2+4=4\left(\frac{1}{16}\right)-17\left(\frac{1}{4}\right)+4=\frac{1}{4}-\frac{17}{4}+\frac{16}{4}=0\end{array}$

Hence,

The zeros of f are $\text{2,-2,}\frac{1}{2}\text{ and }-\frac{1}{2}$