Answer
$P(x)=(x-1)^{2}(x+2)(x-2)$
Zeros: $ \ -2,\ 1, \ 2 \ $
Work Step by Step
$P(x)=x^{4}-2x^{3}-3x^{2}+8x-4$
$P(-x)=x^{4}+2x^{3}-3x^{2}-8x-4$
Descart's rule of signs:
P(x) has 3 sign changes $\Rightarrow$ 3 or 1 positive zeros.
P(-x) has 1 sign changes $\Rightarrow$ 1 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4$
Testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & -2 & -3 & 8 & -4\\\hline
& 1 & -1 & -4 & 4\\\hline
1& -1 & -4 & 4&|\ \ 0\end{array}$
$P(x)=(x-1)(x^{3}-x^{2}-4x+4)$
Again, testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1& -1 & -4 & 4\\\hline
& 1 & 0 & -4\\\hline
1& 0 & -4 &|\ \ 0\end{array}$
$P(x)=(x-1)^{2}(x^{2}-4)$
... recognize a difference of squares
$P(x)=(x-1)^{2}(x+2)(x-2)$
Zeros: $ \ -2,\ 1, \ 2 \ $
