Answer
Zeros:$\displaystyle \quad\pm 2,3,\frac{1}{3}$
$P(x)= (3x-1)(x-2)(x+2)(x-3)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llllll}
p: & \pm 1 & \pm 2, & \pm 3, & \pm 6, & \pm 12\\
q: & \pm 1 & \pm 3 & & &
\end{array}\right.$
$P(x)$ has 3 sign changes, we expect 3 or 1 positive real zeros.
With synthetic division, we try the integers first. 1 is not a zero ... try 2:
\begin{array}{rrrrrrr}
2 \ \rceil & 3 & -10 & -9 & 40 & -12 & \\
& & 6 & -8 & -34 & 12 & \\
\hline & 3 & -4 & -17 & 6 & 0 & \\
& & & & & & \\
3 \ \rceil & 3 & -4 & -17 & 6 & & \\
& & 9 & 15 & -6 & & \\
\hline & 3 & 5 & -2 & 0 & & \\
& & & & & & \\
-2 \ \rceil & 3 & 5 & -2 & & & \\
& & -6 & 2 & & & \\
\hline & 3 & -1 & 0 & & &P(x)= (3x-1)(x-2)(x+2)(x-3) \\
& & & & & & \\
1/3 \ \rceil & 3 & -1 & & & & \\
& & 1 & & & & \\
\hline & 3 & 0 & & & & \\
\end{array}
Zeros:$\displaystyle \quad\pm 2,3,\frac{1}{3}$
P(x)= (3x-1)(x-2)(x+2)(x-3)
