## Precalculus (6th Edition) Blitzer

The zeros of $f\left( x \right)=-{{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{2}}$ are $2,2,-1,-1$.
Let’s first put $f\left( x \right)=0$. Then, \begin{align} & -{{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{2}}=0 \\ & {{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{2}}=0 \end{align} Now, factorize the above equation written as: $\left( x-2 \right)\left( x-2 \right)\left( x+1 \right)\left( x+1 \right)=0$ 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, $\left( x+1 \right)=0$ So, the zeros of the provided function are $2,2,-1,-1$. When $x=0$ $-{{\left( 0-2 \right)}^{2}}{{\left( 0+1 \right)}^{2}}=-4$ The zeros of the graph $2,-1$ have multiplicities 2; the graph touches the x- axis and turns around. The degree of the function is 4 and the leading coefficient is $-1$.