Answer
The zeros of the given function are:
$\color{blue}{\left\{i, -i, -1, 1\right\}}$
The completely factored form of the given function is:
$\color{blue}{P(x) = (x-1)(x+1)(x-i)(x+i)}$
Work Step by Step
RECALL:
$a^2-b^2=(a-b)(a+b)$
The given polynomial function can be written as $P(x) = (x^2)^2-1^2$.
Factor the difference of two squares using the formula above with $a=x^2$ and $b=1$ to obtain:
$P(x) = (x^2+1)(x^2-1)$
Factor the second binomial using the same formula above to obtain:
$P(x) = (x^2+1)(x-1)(x+1)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccccc}
&x^2+1=0 &\text{or} &x-1=0 &\text{or} &x+1=0
\\&x^2=-1 &\text{or} &x=1 &\text{or} &x=-1
\\&x=\pm\sqrt{-1} &\text{or} &x=1 &\text{or} &x=-1
\\&x=\pm i &\text{or} &x=1 &\text{or} &x=-1
\end{array}
Thus, the zeros of the given function are:
$\color{blue}{\left\{i, -i, -1, 1\right\}}$
The completely factored form of the given function is:
$\color{blue}{P(x) = (x-1)(x+1)(x-i)(x+i)}$
