Answer
$P(x)=(2x-3i)(2x+3i)$
The zeros of the function are: $\{\frac{3i}{2}, \frac{-3i}{2}\}$
$x = \frac{3i}{2}$ with multiplicity 1
$x = \frac{-3i}{2}$ with multiplicity 1
Work Step by Step
Factor the polynomial completely to obtain:
$P(x)=4x^{2}+9$
$P(x)=(2x-3i)(2x+3i)$
Equate each unique factor to zero then solve each equation to obtain:
$2x-3i=0 \rightarrow x=\frac{3i}{2}$
$2x+3i=0 \rightarrow x=\frac{-3i}{2}$
The zeros of the function are: $\{\frac{3i}{2}, \frac{-3i}{2}\}$
$x = \frac{3i}{2}$ with multiplicity 1
$x = \frac{-3i}{2}$ with multiplicity 1
