Answer
$P(x)=(x-4)[x+(2+2\sqrt 3i)][x+(2-2\sqrt 3i)]$
The zeros of the function are: $\{4,-2+2\sqrt 3i,-2-2\sqrt 3i\}$
$x =-2+2\sqrt 3i$ with multiplicity 1
$x =-2-2\sqrt 3i$ with multiplicity 1
$x =4$ with multiplicity 1
Work Step by Step
Factor the polynomial completely to obtain:
$P(x)=x^{3}-64$
$P(x)=(x-4)(x^{2}+4x+16)$
$P(x)=(x-4)[x+(2+2\sqrt 3i)][x+(2-2\sqrt 3i)]$
Equate each unique factor to zero then solve each equation to obtain:
$x-4=0 \rightarrow x=4$
$x+(2+2\sqrt 3i)=0 \rightarrow x=-2-2\sqrt 3i$
$x+(2-2\sqrt 3i)=0 \rightarrow x=-2+2\sqrt 3i$
The zeros of the function are: $\{4,-2+2\sqrt 3i,-2-2\sqrt 3i\}$
$x =-2+2\sqrt 3i$ with multiplicity 1
$x =-2-2\sqrt 3i$ with multiplicity 1
$x =4$ with multiplicity 1
