Answer
(a) $\left\{-3i, 3i\right\}$.
(b) $P(x) = (x-3i)^2(x+3i)^2$
Work Step by Step
$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = (x^2+3)(x^2+3)
\\P(x) = [x-3i)(x+3i)] \cdot [(x-3i)(x+3i)]
\\P(x) = (x-3i)^2(x+3i)^2$
Equate each unique factor to zero then solve each equation to obtain:
\begin{array}{ccc}
&x-3i=0 &\text{or} &x+3i=0
\\&x=3i &\text{or} &x=-3i &
\end{array}
Thus, the zeros of the function are: $\left\{-3i, 3i\right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = (x-3i)^2(x+3i)^2$
