Chapter 4.7, Problem 50WE

To determine

To find zero of f.

Answer to Problem 50WE

The zeros of f are $-3\text{ and 3}$

Explanation of Solution

Given information:

  $f\left(y\right)={\left(y^2-9\right)}^3$

Calculation:

The value of f can be calculated as

  $\begin{array}{l}f\left(y\right)={\left(y^2-9\right)}^3\\ {\left(y^2-9\right)}^3=0\left[\text{write }f\left(y\right)=0\right]=\\ \left(y^2-3^2\right)\left(y^2-3^2\right)\left(y^2-3^2\right)\text{=0 }\left[\text{difference of squares}\right]\\ \left(y+3\right)\left(y-3\right)\left(y+3\right)\left(y-3\right)\left(y+3\right)\left(y-3\right)=0\end{array}$

Therefore,

  $\begin{array}{l}\left(y+3\right)=0\text{ or }\left(y-3\right)=0\left[\text{use the zero-product property}\right]\\ y=-3\text{ or }y=3\\ \\ \text{check: }f\left(3\right)={\left(3^2-9\right)}^3={\left(9-9\right)}^3={\left(0\right)}^3=0\\ f\left(-3\right)={\left({\left(-3\right)}^2-9\right)}^3={\left(9-9\right)}^3={\left(0\right)}^3=0\\ \end{array}$

Hence,

The zeros of f are $-3\text{ and 3}$