Answer
Zeros:$\displaystyle \quad 2,\quad \frac{1}{3},\quad- \frac{1}{4}$
$P(x)=(3x-1)(4x+1)(x-2)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{lllllll}
p: & \pm 1 & \pm 2 & & & & \\
q: & \pm 1 & \pm 2, & \pm 3, & \pm 4, & \pm 6 & \pm 12
\end{array}\right.$
$P(x)$ has 2 sign changes, we expect 2 or 0 positive real zeros.
$P(-x)=-12x^{3}-25x^{2}-x+3$
has 1 sign change, so we expect 1 negative zero.
With synthetic division, we try the integers first. 1 is not a zero,...try 2:
\begin{array}{rrrrr}
2 \ \rceil & 12 & -25 & 1 & 2 \\
& & 24 & -2 & -2 \\
\hline & 12 & -1 & -1 & 0 \\
& & & & \\
1/3 \ \rceil & 12 & -1 & -1 & \\
& & 4 & 1 & \\
\hline & 12 & 3 & 0 & \\
& & & & \\
- 1/4 \ \rceil & 12 & 3 & & \\
& & -3 & & \\
\hline & 12 & 0 & & \\
\end{array}
Zeros:$\displaystyle \quad 2,\quad \frac{1}{3},\quad- \frac{1}{4}$
$P(x)=12(x-\displaystyle \frac{1}{3})(x+\frac{1}{4})(x-2)$
$=3(x-\displaystyle \frac{1}{3})\cdot 4(x+\frac{1}{4})(x-2)$
$=(3x-1)(4x+1)(x-2)$