Answer
$P(x)=x(x-i\sqrt{3})^{2}(x+i\sqrt{3})^{2}$
Zeros:
$0$ (multiplicity 1)
$\pm i\sqrt{3}$ (each multiplicity 2)
Work Step by Step
Factor out $x$,
$P(x)=x(x^{4}+6x^{2}+9)$
The parentheses hold
the square of $x^{2}=x^{4},$
the square of $3=9$, and
$2\times(x^{2})\times(3)=6x^{2}$, so it is a square of a sum,
$P(x)=x(x^{2}+3)^{2}$
The expression $x^{2}+3$ is factored as a difference of squares:
$x^{2}+3=x^{2}-3(-1)=x^{2}-(i\sqrt{3})^{2}$
$=(x-i\sqrt{3})(x+i\sqrt{3})$
$P(x)=x[(x-i\sqrt{3})(x+i\sqrt{3})]^{2}$
$P(x)=x(x-i\sqrt{3})^{2}(x+i\sqrt{3})^{2}$
Zeros:
$0$ (multiplicity 1)
$\pm i\sqrt{3}$ (each multiplicity 2)
