Answer
$1,3,\frac{-1\pm\sqrt 7i}{2}$ each has a multiplicity of 1.
Work Step by Step
$P(x)=6x^3(x-3)+6x^2-18x-12x+36=6x^3(x-3)+6x(x-3)-12(x-3)
=6(x-3)(x^3+x-2)$
Clearly we can identify two zeros as $3,1$
$P(x)=6(x-3)(x-1)(x^2+x+2)$
4 zeros $1,3,\frac{-1\pm\sqrt 7i}{2}$ each has a multiplicity of 1.
