Answer
Zeros:$\displaystyle \quad \frac{1}{2},\quad \frac{3}{4},\quad- \frac{2}{3}$
$P(x)=(2x-1)(4x-3)(3x+2)$
Work Step by Step
Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llllllll}
p: & \pm 1 & \pm 2 & \pm 3, & \pm 6 & & & \\
q: & \pm 1 & \pm 2, & \pm 3, & \pm 4, & \pm 6 & \pm 12 & \pm 24
\end{array}\right.$
$P(x)$ has 1 sign change, so we expect 1 positive real zero.
$P(-x)=-24x^{3}+10x^{2}+13x-6$
has 2 sign changes, we expect 2 or 0 negative zeros.
With synthetic division, we try the integers first.
None of the integers are zeros. Try the fractions ... $\pm$1/2...
\begin{array}{rrrrr}
- 1/2\ \rceil & 24 & 10 & -13 & -6 \\
& & -12 & 1 & 6 \\
\hline & 24 & -2 & -12 & 0 \\
& & & & \\
- 2/3 \ \rceil & 24 & -2 & -12 & \\
& & -16 & 12 & \\
\hline & 24 & -18 & 0 & \\
& & & & \\
3/4 \ \rceil & 24 & -18 & & \\
& & 18 & & \\
\hline & 24 & 0 & & \\
\end{array}
Zeros:$\displaystyle \quad \frac{1}{2},\quad \frac{3}{4},\quad- \frac{2}{3}$
$P(x)=24(x-\displaystyle \frac{1}{2})(x-\frac{3}{4})(x+\frac{2}{3})$
$=2(x-\displaystyle \frac{1}{2})\cdot 4(x-\frac{3}{4})\cdot 3(x+\frac{2}{3})$
$=(2x-1)(4x-3)(3x+2)$
