Answer
$P(x)=(2x-1)(2x+3)(3x-1)$
zeros: $-\displaystyle \frac{3}{2},\frac{1}{3},\frac{1}{2}$
Work Step by Step
Descart's rule of signs:
$P(x)=12x^{3}-20x^{2}+x+3$
... $2$ sign changes: 2 or 0 positive zeros
$P(-x)=-12x^{3}-20x^{2}-x+3$
... $1$ sign changes: 1 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,3}{1,2,3,4,6,12}$
Testing with synthetic division, try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
12 &-20 & 1 & 3 & & & \\\hline
& 12 & 8 & 9 & & & \\\hline
12& 8 & 9 & |\ \ 12 & & & \end{array}$
$1$ is not a zero, but it is an upper bound (bottom row all positive)
try 1/2
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
12 &-20 & 1 & 3 & & & \\\hline
& 6 & -7 & -3 & & & \\\hline
12& -14 & -6 & |\ \ 0 & & & \end{array}$
$P(x)=(x-\displaystyle \frac{1}{2})(12x^{2}-14x-6)$
$=(x-\displaystyle \frac{1}{2})\cdot 2(6x^{2}-7x-3)$
$=(2x-1)(6x^{2}-7x-3)$
... searching for factors of $6$($-3$)=$-18$ whose sum is $-7$
... we find $-9$ and $+2$.
$6x^{2}-7x-3=6x^{2}+9x-2x-3$
$=3x(2x+3)-(2x+3)$
$=(2x+3)(3x-1)$
$P(x)=(2x-1)(2x+3)(3x-1)$
zeros: $-\displaystyle \frac{3}{2},\frac{1}{3},\frac{1}{2}$
