Chapter 4.7, Problem 46WE

To determine

To find zero of f.

Answer to Problem 46WE

The zeros of f are $1,2\text{ and 3}$

Explanation of Solution

Given information:

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

Calculation:

The value of f can be calculated as

  $\begin{array}{l}f\left(t\right)={\left(t-2\right)}^3-\left(t-2\right)\\ {\left(t-2\right)}^3-\left(t-2\right)=0\left[\text{write }f\left(t\right)=0\right]\\ \left(t-2\right)\left[{\left(t-2\right)}^2-1\right]=0\left[\text{take out the}\text{common factor,}\left(t-2\right)\right]\\ \left(t-2\right)\left[{\left(t-2\right)}^2-{\left(1\right)}^2\right]=0\left[\text{difference of squares}\right]\\ \left(t-2\right)\left[\left(t-2+1\right)\left(t-2-1\right)\right]=0\\ \left(t-2\right)\left(t-1\right)\left(x-3\right)=0\left[\text{simplify}\right]\end{array}$

Therefore,

  $\begin{array}{l}\left(t-1\right)\text{=0 or }\left(t-2\right)=0\text{ or }\left(t-3\right)\text{=0 }\left[\text{use the zero-product property}\right]\\ t=1\text{ or t=2 or t}=3\\ \\ \text{check: }f\left(1\right)={\left(1-2\right)}^3-\left(1-2\right)={\left(-1\right)}^3-\left(-1\right)=-1+1=0\\ f\left(2\right)={\left(2-2\right)}^3-\left(2-2\right)={\left(0\right)}^3-\left(0\right)=0-0=0\\ f\left(3\right)={\left(3-2\right)}^3-\left(3-2\right)={\left(1\right)}^3-\left(1\right)=1-1=0\end{array}$

Hence,

The zeros of f are $1,2\text{ and 3}$