## Precalculus (6th Edition) Blitzer

The zeros are $2,2,-1,-1,-1$.
First put $f\left( x \right)=0$. So, ${{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{3}}=0$ Now, factorize the above equation written as: $\left( x-2 \right)\left( x-2 \right)\left( x+1 \right)\left( x+1 \right)\left( x+1 \right)=0$ Put each factor equal to $0$. So, $\left( x-2 \right)=0$ Or, $\left( x-2 \right)=0$ Or, $\left( x+1 \right)=0$ Or, $\left( x+1 \right)=0$ Or, \begin{align} & \left( x+1 \right)=0 \\ & \Rightarrow x=2,2,-1,-1,-1 \end{align} So, the zeros of the provided function are $2,2,-1,-1,-1$. Graph of the function $f\left( x \right)={{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{3}}$ shown below: