Answer
zeros:$\displaystyle \qquad -2,\ 1,\ \frac{-1\pm\sqrt{5}}{2}$
Work Step by Step
$P(x)=x^{4}+2x^{3}-2x^{2}-3x+2,$
($2$ sign variations, we expect $2$ or $0$ real zero)
$P(-x)=x^{4}-2x^{3}-2x^{2}+3x+2,$
($2$ sign variations, we expect $2$ or $0$ negative real zeros)
$a_{0}=2 \quad $p: $\pm 1,\pm 2$
$a_{n}=1,\qquad $q: $\pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
1 & 1 & 2 & -2 & -3 & 2 & & \\
& & 1 & 3 & 1 & -2 & & \\
\hline & 1 & 3 & 1 & -2 & 0 & & \\
& & & & & & & \\
-2 & 1 & 3 & 1 & -2 & & & \\
& & -2 & -2 & 2 & & & \\
\hline & 1 & 1 & -1 & 0 & & & \\
\end{array}$
$P(x)=(x-1)(x+2)(x^{2}+x-1)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{-1\pm\sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}=\frac{-1\pm\sqrt{5}}{2}$
zeros:$\displaystyle \qquad 1,\ -2,\ \frac{-1\pm\sqrt{5}}{2}$