Answer
The zeros of the given function are:
$\color{blue}{\left\{i, -i, -2i, 2i\right\}}$
The completely factored form of the given function is:
$\color{blue}{P(x) = (x-i)(x+i)(x-2i)(x+2i)}$
Work Step by Step
Factor the trinomial to obtain:
$P(x) = (x^2+4)(x^2+1)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccc}
&x^2+4=0 &\text{or} &x^2+1=0
\\&x^2=-4 &\text{or} &x^2=-1
\\&x=\pm\sqrt{-4} &\text{or} &x=\pm\sqrt{-1}
\\&x=\pm\sqrt{4(-1)} &\text{or} &x=\pm i
\\&x=\pm\sqrt{2^2(-1)} &\text{or} &x=\pm i
\\&x=\pm2\sqrt{-1} &\text{or} &x=\pm i
\\&x=\pm2i &\text{or} &x=\pm i
\end{array}
Thus, the zeros of the given function are:
$\color{blue}{\left\{i, -i, -2i, 2i\right\}}$
The completely factored form of the given function is:
$\color{blue}{P(x) = (x-i)(x+i)(x-2i)(x+2i)}$
