Answer
$x=0$ with a multiplicity of $1$
$x=-5$ with a multiplicity of $1$
$x=\pm4$ each with a multiplicity of $1$
Work Step by Step
Step 1. Give $f(x)=5x^2(x^2-16)(x+5)$, we can find the first zero as $x=0$ with a multiplicity of $1$
Step 2. We can find the second zero as $x=-5$ with a multiplicity of $1$
Step 3. Let $(x^2-16)=0$, we have $x=\pm4$ each with a multiplicity of $1$