Answer
$(a)\quad $Zeros: $0$ and $\pm 3i$
$(b)\quad P(x)=x^{3}(x-3i)(x+3i)$
Work Step by Step
Complete Factorization Theorem (p.287)
$P$ factors into $n$ linear factors : $P(x)=a(x-c_{1})(x-c_{2})\cdots(x-c_{n})$
where $a$ is the leading coefficient of $P$ and $c_{1}, c_{1}, \ldots, c_{n}$ are the zeros of $P$.
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$(a)\quad$
$x^{5}+9x^{3}=0$
$x^{3}(x^{2}+9)=0$
So,
either
$x^{3}=0\Rightarrow x=0$ is a zero, (multiplicity 3),
or
$x^{2}+9=0\Rightarrow x^{2}=-3\Rightarrow x=\pm 3i$
Zeros: $0$ and $\pm 3i$
$(b)$
The leading coefficient of P is $a=1$, so, by the above theorem,
$P(x)=x^{3}(x-3i)(x+3i)$