Answer
Zeros:$\displaystyle \quad -1,\quad \pm\frac{1}{2}$
$P(x)=(2x-1)(2x+1)(x+1)$
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.
$P(-x)=-x^{3}+4x^{2}+x-1$
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}{rrrrr}
-1 \ \rceil & 4 & 4 & -1 & -1 \\
& & -4 & 0 & 1 \\
\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:$\displaystyle \quad -1,\quad \pm\frac{1}{2}$
$P(x)=4(x-\displaystyle \frac{1}{2})(x+\frac{1}{2})(x+1)$
$=2(x-\displaystyle \frac{1}{2})\cdot 2(x+\frac{1}{2})(x+1)$
$=(2x-1)(2x+1)(x+1)$
