Answer
$0$ is a zero of multiplicity $2$.
$-5$ is a zero of multiplicity $1.$
$4$ is a zero of multiplicity $1.$
$-4$ is a zero of multiplicity $1.$
Work Step by Step
$x=k$ is a zero of a polynomial function $f(x)$, then $f(k)=0..$
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
The number of times $(x-k) $ occurs as a factor is referred to as the multiplicity of the zero.
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$x^{2}=(x-0)^{2}$ is a factor of f $\Rightarrow k_{1}=0$ is a zero of multiplicity 2.
$(x+5)$ is a factor of f $\Rightarrow k_{2}=-5$ is a zero of multiplicity $1.$
$x^{2}-16$ is a factor of f. Its zeros are $\pm 4 \quad \left(\begin{array}{l}
x^{2}-16=0\\
x^{2}=16\\
x=\pm 4
\end{array}\right).$
Thus, $(x-4)$ and $(x+4)$ are factors of f, so
$k_{3}=4$ is a zero of multiplicity $1.$
$k_{4}=-4$ is a zero of multiplicity $1.$
