Answer
Zeros: $-1, 2, 1/2.$
$P(x)=(x+1)(x-2)^2(2x-1)$
Work Step by Step
$P(x)=2x^{4}-7x^{3}+3x^{2}+8x-4,$
($3$ sign variations, we expect $3$ or $1$ real zeros)
$P(-x)=2x^{4}+7x^{3}+3x^{2}-8x-4,$
($1$ sign variations, we expect $1$ negative real zero)
$a_{0}=-4, \quad $p: $\pm 1,\pm 2,\pm 4$
$a_{n}=12,\qquad $q: $\pm 1.,\pm 2,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1& 2 & -7 & 3 & 8 & -4 & & \\
& & -2 & 9 & -12 & 4 & & \\
\hline & 2 & -9 & 12 & -4 & 0 & & \\
& & & & & & P(x)&=(x+1)(2x^{3}-9x^{2}+12x-4) \\\\
2& 2 & -9 & 12 & -4 & & & \\
& & 4 & -10 & 4 & & & \\
\hline & 2 & -5 & 2 & 0 & & & \\
& & & & & & P(x)&=(x+1)(x-2)(2x^{2}-5x+2) \\\\
2 & 2 & -5 & 2 & & & & \\
& & 4 & -2 & & & & \\
\hline & 2 & -1 & 0 & & & & \\
1/2 & & & & & & P(x)&=(x+1)(x-2)(x-2)(2x-1) \\\\
& 2 & -1 & & & & & \\
& & 1 & & & & & \\
\hline & 2 & 0 & & & & & \\
\end{array}$
Zeros: -$1, 2, 1/2.$
$P(x)=(x+1)(x-2)^{2}(x-2)(x-1/2)\cdot 2$
$=(x+1)(x-2)(x-2)(2x-1)$
