Answer
$P(x)=(x-1)^{2}(x+2)(x-2)$
Zeros: $\pm 2, 1$
Work Step by Step
$P(x)=x^{4}-2x^{3}-3x^{2}+8x-4,$
($3$ sign variation, we expect $1$ or $3$ positive real zero)
$P(-x)=x^{4}+2x^{3}-3x^{2}-8x-4,$
($1$ sign variations, we expect $1$ negative real zeros)
$a_{0}=-4, \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:
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & -2 & -3 & 8 & -4 & \\
& & 1 & -1 & -4 & 4 & \\
& -- & -- & -- & -- & -- & \\
& 1 & -1 & -4 & 4 & \underline{|0} & \Rightarrow P(1)=0,
\end{array}\right.$
$P(x)=(x-1)(x^{3}-x^{2}-4x+4)$
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 ...
$=(x-1)(x+2)(x-2)$
$P(x)=(x-1)(x-1)(x+2)(x-2)$
$P(x)=(x-1)^{2}(x+2)(x-2)$
Zeros: $\pm 2, 1$