Answer
$P(x)=(4x-3)(2x+1)(3x+2)$
zeros: $-\displaystyle \frac{1}{2}, -\frac{2}{3}, \frac{3}{4}$
Work Step by Step
Descart's rule of signs:
$P(x)=24x^{3}+10x^{2}-13x-6$
... $1$ sign changes: 1 positive zeros
$P(-x)=-24x^{3}+10x^{2}+13x-6$
... $2$ sign changes: 2 or 0 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,3,6}{1,2,3,4,6,8,12,24}$
Testing with synthetic division, try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
24 &10 &-13 & -6 & & & \\\hline
& 24 & 34 & 21 & & & \\\hline
24&34 & 21 & |\ \ 15 & & & \end{array}$
$1$ is not a zero, but it is an upper bound (bottom row all positive)
trying 1/2, 1/3, 2/3 all fail. Next up, 3/4:
$\left.\begin{array}{l}
3/4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
24 &10 &-13 & -6 & & & \\\hline
& 18 & 21 & 6 & & & \\\hline
24&28 & 8 & |\ \ 0 & & & \end{array}$
$P(x)=(x-\displaystyle \frac{3}{4})(24x^{2}+28x+8)$
$=(x-\displaystyle \frac{3}{4})\cdot 4(6x^{2}+7x+2)$
$=(4x-3)(6x^{2}+7x+2)$
... searching for factors of 6(2)=12 whose sum is 7
... we find 3 and 4.
$6x^{2}+7x+2=6x^{2}+3x+4x+2$
$=3x(2x+1)+2(2x+1)$
$=(2x+1)(3x+2)$
$P(x)=(4x-3)(2x+1)(3x+2)$
zeros: $-\displaystyle \frac{1}{2}, -\frac{2}{3}, \frac{3}{4}$
