Answer
$P(x)=(x-2)(3x-1)(4x+1)$
zeros: $-\displaystyle \frac{1}{4}, \frac{1}{3}, 2$
Work Step by Step
Descart's rule of signs:
$P(x)=12x^{3}-25x^{2}+x+2$
... $2$ sign changes: 2 or 0 positive zeros
$P(-x)=-12x^{3}-25x^{2}-x+2$
... 1 sign changes: 1 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2}{1,2,3,4,6,12}$
Testing with synthetic division, try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
12 &-25 &1 & 2 & & & \\\hline
& 12 & -13 & -12 & & & \\\hline
12 &-13 &-12 & |\ \ -10 & & & \end{array}$
x=1 fails, but it is not an upper bound. Try 2
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
12 &-25 &1 & 2 & & & \\\hline
& 24 & -2 & -2 & & & \\\hline
12 &-1 &-1 & |\ \ -0 & & & \end{array}$
$P(x)=(x-2)(12x^{2}-x-1)$
... we find factors of 12(-1)=-12 whose sum is -1
... they are -4 and 3
$12x^{2}-x-1=12x^{2}-4x+3x-1$
$=4x(3x-1)+(3x-1)$
$=(3x-1)(4x+1)$
$P(x)=(x-2)(3x-1)(4x+1)$
zeros: $-\displaystyle \frac{1}{4}, \frac{1}{3}, 2$
