Answer
$x=3$ with a multiplicity of $3$
$x=\frac{1}{2}$ with a multiplicity of $3$
$x=2+\sqrt 5$ each with a multiplicity of $1$
Work Step by Step
Step 1. Give $f(x)=(2x^2-7x+3)^3(x-2-\sqrt 5)=(x-3)^3(2x-1)^3(x-2-\sqrt 5)$, we can find the first zero as $x=3$ with a multiplicity of $3$
Step 2. We can find the second zero as $x=\frac{1}{2}$ with a multiplicity of $3$
Step 3. Let $x-2-\sqrt 5=0$, we have $x=2+\sqrt 5$ each with a multiplicity of $1$
