Answer
Zeros:$\displaystyle \quad 1,\quad \frac{1}{2},\quad- \frac{3}{2}$
$P(x)=(2x-1)(1x+3)(x-1)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{lll}
p: & \pm 1 & \pm 3\\
q: & \pm 1 & \pm 2
\end{array}\right.$
$P(x)$ has 2 sign change2, we expect 2 or 0 positive real zero.
$P(-x)=-4x^{3}+7x+3$
has 1 sign change, so we expect 1 negative zero.
With synthetic division, we try the integers first. try 1:
\begin{array}{rrrrr}
1 \ \rceil & 4 & 0 & -7 & 3 \\
& & 4 & 4 & -3 \\
\hline & 4 & 4 & -3 & 0 \\
& & & & \\
1/2\ \rceil & 4 & 4 & -3 & \\
& & 2 & 3 & \\
\hline & 4 & 6 & 0 & \\
& & & & \\
-3/2 \ \rceil & 4 & 6 & & \\
& & -6 & & \\
\hline & 4 & 0 & & \\
\end{array}
Zeros:$\displaystyle \quad 1,\quad \frac{1}{2},\quad- \frac{3}{2}$
$P(x)=4(x-\displaystyle \frac{1}{2})(x+\frac{3}{2})(x-1)$
$=2(x-\displaystyle \frac{1}{2})\cdot 2(x+\frac{3}{2})(x-1)$
$=(2x-1)(1x+3)(x-1)$
