Answer
$P(x)=(x+1)(x-2)(2x-1)(3x+1)$
zeros: $-1, -1/3, 1/2, 2$
Work Step by Step
$P(x)=6x^{4}-7x^{3}-12x^{2}+3x+2,$
($2$ sign variations, we expect $2$ or $0$ real zeros)
$P(-x)=6x^{4}+7x^{3}-12x^{2}-3x+2,$
(2 sign variations, we expect 2 or 0 negative real zero)
$a_{0}=2, \quad $p: $\pm 1,\pm 2,$
$a_{n}=6,\qquad $q: $\pm 1.,\pm 2,\pm 3,\pm 6$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 6 & -7 & -12 & 3 & 2 & & \\
& & -6 & 13 & -1 & -2 & & \\
\hline & 6 & -13 & 1 & 2 & 0 & & \\
& & & & & & P(x) &=(x+1)(6x^{3}-13x^{2}+x+2) \\\\
2 & 6 & -13 & 1 & 2 & & & \\
& & 12 & -2 & -2 & & & \\
\hline & 6 & -1 & -1 & 0 & & & \\
& & & & & & P(x) &=(x+1)(x-2)(6x^{2}-x-1) \\\\
1/2 & 6 & -1 & -1 & & & & \\
& & 3 & 1 & & & & \\
\hline & 6 & 2 & 0 & & & & \\
& & & & & & P(x) &=(x+1)(x-2)(x-\displaystyle \frac{1}{2})(6x-2) \\\\
- 1/3 & 6 & 2 & & & & & \\
& & -2 & & & & & \\
\hline & 6 & 0 & & & & &
\end{array}$
$ P(x) =(x+1)(x-2)(x-\displaystyle \frac{1}{2})(x+\frac{1}{3})\cdot 6$
$P(x)= (x+1)(x-2)\displaystyle \cdot 2(x-\frac{1}{2})\cdot 3(x+\frac{1}{3})$
$P(x)=(x+1)(x-2)(2x-1)(3x+1)$
zeros: $-1, -1/3, 1/2, 2$