Answer
$P(x)=(2x-3)(2x+3)(2x+3i)(2x-3i)$
The zeros of the function are: $\{\frac{3}{2},\frac{-3}{2},\frac{3i}{2},\frac{-3i}{2}\}$
$x =\frac{-3}{2}$ with multiplicity 1
$x =\frac{3}{2}$ with multiplicity 1
$x =\frac{-3i}{2}$ with multiplicity 1
$x =\frac{3i}{2}$ with multiplicity 1
Work Step by Step
Factor the polynomial completely to obtain:
$P(x)=16x^{4}-81$
$P(x)=(4x^{2}-9)(4x^{2}+9)$
$P(x)=(2x-3)(2x+3)(2x+3i)(2x-3i)$
Equate each unique factor to zero then solve each equation to obtain:
$2x-3=0 \rightarrow x=\frac{3}{2}$
$2x+3=0 \rightarrow x=\frac{-3}{2}$
$2x+3i=0 \rightarrow x=\frac{-3i}{2}$
$2x-3i=0 \rightarrow x=\frac{3i}{2}$
The zeros of the function are: $\{\frac{3}{2},\frac{-3}{2},\frac{3i}{2},\frac{-3i}{2}\}$
$x =\frac{-3}{2}$ with multiplicity 1
$x =\frac{3}{2}$ with multiplicity 1
$x =\frac{-3i}{2}$ with multiplicity 1
$x =\frac{3i}{2}$ with multiplicity 1
