Answer
Zeros:$ \quad \displaystyle \pm 3,\quad\pm\frac{1}{2}$
$P(x)=(2x+1)(2x-1)(x+3)(x-3)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llll}
p: & \pm 1 & \pm 3, & \pm 9\\
q: & \pm 1 & \pm 2 &
\end{array}\right.$
$P(x)$ 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 ... next up, 3:
$\begin{array}{rrrrrr}
3 \rceil & 4 & 0 & -37 & 0 & 9 \\
& & 12 & 36 & -3 & -9 \\
\hline & 4 & 12 & -1 & -3 & 0 \\
& & & & & \\
& & & & & \\
-3 \rceil & 4 & 12 & -1 & -3 & \\
& & -12 & 0 & 3 & \\
\hline & 4 & 0 & -1 & 0 & \\
& & & & & \\
& & & & & \\
1/2 \rceil & 4 & 0 & -1 & & \\
& & 2 & 1 & & \\
\hline & 4 & 2 & 0 & & \\
& & & & & \\
& & & & & \\
- 1/2 \rceil & 4 & 2 & & & \\
& & -2 & & & \\
\hline & 4 & 0 & & & \\
\end{array}$
Zeros:$ \quad \displaystyle \pm 3,\quad\pm\frac{1}{2}$
$P(x)=4(x+\displaystyle \frac{1}{2})(x-\frac{1}{2})(x+3)(x-3)$
$=2(x+\displaystyle \frac{1}{2})\cdot 2(x-\frac{1}{2})(x+3)(x-3)$
$=(2x+1)(2x-1)(x+3)(x-3)$
