Answer
(a) $\left\{0, -2i, 2i\right\}$.
(b) $P(x) = x^2(x-2i)(x+2i)$
Work Step by Step
$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = x^2(x^2+4)
\\P(x) = x^2(x-2i)(x+2i)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccccc}
&x^2=0 &\text{or} &x-2i=0 &\text{or} &x+2i=0
\\&x=0 &\text{or} &x=2i &\text{or} &x=-2i
\end{array}
Thus, the zeros of the function are: $\left\{0, -2i, 2i\right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = x^2(x-2i)(x+2i)$
