Answer
$P(x)=(x+1)(x-1)(x-2i)(x+2i)$
Zeros: $\pm 2i, \pm 1$ (all with multiplicity 1)
Work Step by Step
Let $t=x^{2}$
$t^{2}+3t-4=$
... find factors of $-4$ whose sum is $3.$
... (these are $+4$ and $-1$)
$=(t+4)(t-1)$
... back-substitute:
= $(x^{2}+4)(x^{2}-1)$
First parentheses:
$x^{2}+4=x^{2}+4(-1)=x^{2}-(2i)^{2}$
$=(x-2i)(x+2i)$
Second parentheses:
$x^{2}-1=(x+1)(x-1)$
$P(x)=(x+1)(x-1)(x-2i)(x+2i)$
Zeros: $\pm 2i, \pm 1$ (all with multiplicity 1)
