Answer
$x=0$ with a multiplicity of $1$
$x=2$ with a multiplicity of $1$
$x=-3$ with a multiplicity of $1$
$x=\pm1$ each with a multiplicity of $1$
Work Step by Step
Step 1. Give $f(x)=3x(x-2)(x+3)(x^2-1)$, we can find the first zero as $x=0$ with a multiplicity of $1$
Step 2. We can find the second zero as $x=2$ with a multiplicity of $1$
Step 3. We can find the third zero as $x=-3$ with a multiplicity of $1$
Step 4. Let $(x^2-1)=0$, we have $x=\pm1$ each with a multiplicity of $1$