Answer
The zeros of $f\left( x \right)=-{{\left( x-2 \right)}^{2}}{{\left( x+1 \right)}^{2}}$ are $2,2,-1,-1$.
Work Step by Step
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 $-1$.
