Answer
Zeros: $\displaystyle \frac{1}{2},\frac{3}{2}$ and $-\displaystyle \frac{1}{3}.\quad P(x)=(2x-1)(2x-3)(3x+1)$
Work Step by Step
$P(x)=12x^{3}-20x^{2}+x+3,$
($2$ sign variations, we expect 2 or 0 real zeros)
$P(-x)=-12x^{3}-20x^{2}-x+3,$
($1$ sign variations, we expect $1$ negative real zero)
$a_{0}=3, \quad $p: $\pm 1,\pm 3,$
$a_{n}=12,\qquad $q: $\pm 1.,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12 $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try 1/2, 1, 3/2,...
$\begin{array}{r|rrrrrr}\hline
1/2 & 12 & -20 & 1 & 3 \\
& & 6 & -7 & -3 \\
\hline & 12 & -14 & -6 & 0 \\
& & & & P(x)&=(x-\displaystyle \frac{1}{2})(12x^{2}-14x-6) \\\\
3/2 & 12 & -14 & -6 & \\
& & 18 & 6 & \\
\hline & 12 & 4 & 0 & \\
& & & & P(x)&=(x-\displaystyle \frac{1}{2})(x-\frac{3}{2})(12x+4) \\\\
- 1/3 & 12 & 4 & & \\
& & -4 & & \\
\hline & 12 & 0 & & \\ \hline
\end{array}$
$P(x)=(x-\displaystyle \frac{1}{2})(x-\frac{3}{2})(x+\frac{1}{3})\cdot 12$
$= 2(x-\displaystyle \frac{1}{2})\cdot 2(x-\frac{3}{2})\cdot 3(x+\frac{1}{3})$
$=(2x-1)(2x-3)(3x+1)$
