Answer
Zeros:$\displaystyle \quad -2, \quad \frac{1}{2}$
$P(x)=(2x-1)(x+2)^{2}$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llll}
p: & \pm 1 & \pm 2, & \pm 4\\
q: & \pm 1 & \pm 2 &
\end{array}\right.$
$P(x)$ has $1$ sign change, we expect 1 positive real zero.
$P(-x)=-2x^{3}+7x^{2}-4x-4$
has 2 sign changes, so we expect 2 or 0 negative zeros
With synthetic division, we try the integers first. -1 is not a zero ... try -2:
\begin{array}{rrrrrr}
-2 \ \rceil & 2 & 7 & 4 & -4 & \\
& & -4 & -6 & 4 & \\
\hline & 2 & 3 & -2 & 0 & \\
& & & & & \\
-2 \ \rceil & 2 & 3 & -2 & & \\
& & -4 & 2 & & \\
\hline & 2 & -1 & 0 & & P(x)=(2x-1)(x+2)(x+2) \\
& & & & & \\
1/2\ \rceil & 2 & -1 & & & \\
& & 1 & & & \\
\hline & 2 & 0 & & & \\
\end{array}
Zeros:$\displaystyle \quad -2, \quad \frac{1}{2}$
$P(x)=(2x-1)(x+2)^{2}$
