Answer
$P(x)=(x-1)(x+1)(x+2)(x-2)$
Zeros: $-2, -1, 1$ and $2$
Work Step by Step
$P(x)=x^{4}-5x^{2}+4,$
($2$ sign variation, we expect 0 or 2 positive real zero)
$P(-x)=x^{4}-5x^{2}+4,$
(2 sign variations, we expect 0 or 2 negative real zeros)
$a_{0}=-3, \quad $p: $\pm 1,\pm 2,\pm 4,$
$a_{n}=1,\qquad $q: $\pm 1. $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
$\pm 1,\pm 2,\pm 4.$
Test $x= 1 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & 0 & -5 & 0 & 4 & \\
& & 1 & 1 & -4 & -4 & \\
& -- & -- & -- & -- & -- & \\
& 1 & 1 & -4 & -4 & \underline{|0} & \Rightarrow P(1)=0,
\end{array}\right.$
$P(x)=(x-1)(x^{3}+x^{2}-4x-4)$
To factor the parentheses, group terms two by two,
$x^{3}+x^{2}-4x-4=x^{2}(x+1)-4(x+1)$
$=(x+1)(x^{2}-4)$
... difference of squares ...
$=(x+1)(x+2)(x-2)$
$P(x)=(x-1)(x+1)(x+2)(x-2)$
Zeros: $-2, -1, 1$ and $2$
