Answer
(a) $\left\{0, -3i, 3i\right\}$.
(b) $P(x) = x^3(x-3i)(x+3i)$
Work Step by Step
$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = x^3(x^2+9)
\\P(x) = x^3(x-3i)(x+3i)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccccc}
&x^3=0 &\text{or} &x-3i=0 &\text{or} &x+3i=0
\\&x=0 &\text{or} &x=3i &\text{or} &x=-3i
\end{array}
Thus, the zeros of the function are: $\left\{0, -3i, 3i\right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = x^3(x-3i)(x+3i)$
