Answer
Zeros: $\displaystyle \frac{3}{2}$ , $-\displaystyle \frac{3}{2}$ , $\displaystyle \frac{3}{2}i$ , $-\displaystyle \frac{3}{2}i$,
each with multiplicity 1.
$P(x)=(2x-3)(2x+3)(2x-3i)(2x+3i)$
Work Step by Step
$P(x)=(4x^{2})^{2}-9^{2}= \qquad $...difference of squares...
$=(4x^{2}-9)(4x^{2}+9)$
$4x^{2}-9= (2x)^{2}-3^{2}\quad $...difference of squares...$=(2x-3)(2x+3)$
$4x^{2}+9= (2x)^{2}-(-1)\cdot 3^{2}=(2x)^{2}-(3i)^{2} \qquad $...difference of squares...$=(2x-3i)(2x+3i)$
$P(x)=(2x-3)(2x+3)(2x-3i)(2x+3i)$
$\left[\begin{array}{llll}
2x-3=0 & 2x+3=0 & 2x-3i=0 & 2x+3i=0\\
2x=3 & 2x=-3 & 3x=3i & 2x=-3i\\
x=\frac{3}{2} & x=-\frac{3}{2} & x=\frac{3}{2}i & x=-\frac{3}{2}i
\end{array}\right]$
Zeros: $\displaystyle \frac{3}{2}$ , $-\displaystyle \frac{3}{2}$ , $\displaystyle \frac{3}{2}i$ , $-\displaystyle \frac{3}{2}i$,
each with multiplicity 1.
$P(x)=(2x-3)(2x+3)(2x-3i)(2x+3i)$
