Answer
$P(x)=(x-2)(x-3)(x+2)(3x-1)$
zeros: $-2, \displaystyle \frac{1}{3}, 2, 3$
Work Step by Step
Descart's rule of signs:
$P(x)=3x^{4}-10x^{3}-9x^{2}+40x-12$
... 3 sign changes: 3 or 1 positive zeros
$P(-x)=3x^{4}+10x^{3}-9x^{2}-40x-12$
... 1 sign changes: 1 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,3,4,6,12}{1,3}$
Testing with synthetic division, 1 fails, try 2
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
3 &-10 &-9 & 40 & -12 & & \\\hline
& 6 &-8 & -34 & 12 & & \\\hline
3 &-4 &-17 & 6 & |\ \ 0 & & \end{array}$
$P(x)=(x-2)(3x^{3}-4x^{2}-17x+6)$
Testing with synthetic division again, 2 fails, try 3:
$\left.\begin{array}{l}
3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
3 &-4 & -17 & 6 & & & \\\hline
& 9 &15 & -6 & & & \\\hline
3 &5 &-2 & |\ \ 0 & & & \end{array}$
$P(x)=(x-2)(x-3)(3x^{2}+5x-2)$
We factor the trinomial by searching for factors of 3(-2)=-6 whose sum is 5
... we find 6 and -1
$3x^{2}+5x^{2}-2=3x^{2}+6x-x-2$
$=3x(x+2)-(x+2)=(x+2)(3x-1)$
$P(x)=(x-2)(x-3)(x+2)(3x-1)$
zeros: $-2, \displaystyle \frac{1}{3}, 2, 3$
