Answer
$P(x)=(x+2)^{2}(x-1-\sqrt{3})^{2}(x-1+\sqrt{3})^{2}$
Zeros:
$-2,\ 1\pm i\sqrt{3}$,
all with multiplicity 2.
Work Step by Step
First term is (the square of $x^{3})=x^{6}$
Third term is (the square of 8$)=64$
Second term = $ 2\times$($x^{3})(8)$,
so we have a square of a sum:
$P(x)=(x^{3}+8)^{2}$
The expression $x^{3}+8=x^{3}+2^{3} $is a sum of cubes:
$x^{3}+8=(x+2)(x^{2}-2x+4)$
so
$P(x)=[(x+2)(x^{2}-2x+4)]^{2}$
With the quadratic formula for $x^{2}-2x+4=0,$
$x=\displaystyle \frac{+2\pm\sqrt{4-4(1)(4)}}{2}=\frac{2\pm\sqrt{-12}}{2}$
$=\displaystyle \frac{2\pm i\cdot 2\sqrt{3}}{2}=1\pm i\sqrt{3}$
$x^{2}-2x+4=[x-(1+i\sqrt{3})][x-(1-i\sqrt{3})]$
$=(x-1-\sqrt{3})(x-1+\sqrt{3})$
So
$P(x)=(x+2)^{2}(x-1-\sqrt{3})^{2}(x-1+\sqrt{3})^{2}$
Zeros:
$-2,\ 1\pm i\sqrt{3}$,
all with multiplicity 2.
