Answer
Zeros: $-1,4,-1/2,4/3.$
$P(x)=(3x-4)(2x+1)(x-4)(x+1)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llllll}
p: & \pm 1 & \pm 2, & \pm 4, & \pm 8, & \pm 16\\
q: & \pm 1 & \pm 2, & \pm 3, & \pm 6 &
\end{array}\right.$
$P(x)$ has two sign changes, we expect 2 or 0 positive real zeros.
With synthetic division, we try the integers first. 1 is not a zero ... try $-1$:
\begin{array}{rrrrrr}
-1 \rceil & 6 & -23 & -13 & 32 & 16 \\
& & -6 & 29 & -16 & -16 \\
\hline & 6 & -29 & 16 & 16 & 0 \\
& & & & & \\
4 \rceil & 6 & -29 & 16 & 16 & \\
& & 24 & -20 & -16 & \\
\hline & 6 & -5 & -4 & 0 & \\
& & & & & \\
- 1/2 \rceil & 6 & -5 & -4 & & \\
& & -3 & 4 & & \\
\hline & 6 & -8 & 0 & & \\
& & & & & \\
4/3 \rceil & 6 & -8 & & & \\
& & 8 & & & \\
\hline & 6 & -0 & & & \\
\end{array}
Zeros: $-1,4,-1/2,4/3.$
$P(x)=6(x-\displaystyle \frac{4}{3})(x+\frac{1}{2})(x-4)(x+1)$
$= 3(x-\displaystyle \frac{4}{3})\cdot 2(x+\frac{1}{2})(x-4)(x+1)$
$=(3x-4)(2x+1)(x-4)(x+1)$