Answer
zeros : $-1,-\displaystyle \frac{1}{2}, 4, \frac{4}{3}$
$P(x)=(x+1)(2x+1)(x-4)(3x-4)$
Work Step by Step
Descart's rule of signs:
$P(x)=6x^{4}-23x^{3}-13x^{2}+32x+16$
... 2 sign changes: 2 or 0 positive zeros
$P(-x)=6x^{4}+23x^{3}-13x^{2}-32x+16$
... 2 sign changes: 2 or 0 negative zeros
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4,\pm 8,\pm 16$
candidates for q:$\quad \pm 1,\pm 2,\pm 3,\pm 6$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \displaystyle \pm\frac{1,2,4,8,16}{1,2,3,6}$
Testing with synthetic division,
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
6 &-23 &-13 & 32 & 16 & & \\\hline
& -6 &29 & -16 & -16 & & \\\hline
6 &-29 &16 & 16 & |\ \ 0 & & \end{array}$
$P(x)=(x+1)(6x^{3}-29x^{2}+16x+16)$
Again, testing with synthetic division,
$\left.\begin{array}{l}
-1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
6 &-29 & 16 & 16 & & & \\\hline
& -3 &16 & -16 & & & \\\hline
6 &-32 &32 & |\ \ 0 & & & \end{array}$
$P(x)=(x+1)(x+\displaystyle \frac{1}{2})(6x^{2}-32x+32)$
Again, testing with synthetic division, this time positives. 1 and 2 fail, but
$\left.\begin{array}{l}
4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
6 &-32 & 32 & & & & \\\hline
& 24 &-32 & & & & \\\hline
6 &-8 &|\ \ 0 & & & & \end{array}$
$P(x)=(x+1)(x+\displaystyle \frac{1}{2})(x-4)(6x-8)$
zeros : $-1,-\displaystyle \frac{1}{2}, 4, \frac{4}{3}$
$P(x)=(x+1)(x+\displaystyle \frac{1}{2})(x-4)\cdot 2(3x-4)$
$=(x+1)\displaystyle \cdot 2(x+\frac{1}{2})(x-4)(3x-4)$
$P(x)=(x+1)(2x+1)(x-4)(3x-4)$
