Answer
$P(x)=(x+1)(x-3i)(x+3i).$
Zeros: -1 , $3i$ , $-3i$,
each with multiplicity 1.
Work Step by Step
Factoring in pairs,
$ P(x)=x^{2}(x+1)+9(x+1)=\qquad$ common: $(x+1)$
$P(x)=(x+1)(x^{2}+9)$
$x^{2}+9=x^{2}-(-1)\cdot 3^{2}=x^{2}-(3i)^{2} \qquad $...difference of squares...$=(x-3i)(x+3i)$
$P(x)=(x+1)(x-3i)(x+3i).$
Zeros: -1 , $3i$ , $-3i$,
each with multiplicity 1.