1. ## Solving Equation

solv for x, 6x^3 -7x^2 -12x +7 = 0

2. Using the Rational Zeros Test:
$\displaystyle \dfrac{Factors\f\:7}{Factors\f\:6} = \dfrac{\pm 1,\,\pm 7}{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}$

... and guess-and-check with these potential rational zeros, I see that 1/2 is a zero. So the polynomial factors to:
$\displaystyle 6x^3 - 7x^2 - 12x + 7 = (2x - 1)(3x^2 - 2x - 7)$

The quadratic isn't factorable, so you'll have to use the quadratic formula to get the other two roots.

3. Isn't it?

$\displaystyle (2x - 1)(3x^2 -2x - 7) = 3(2x -1)\left(x^2 - \frac{2}{3}x - \frac{7}{3}\right)$

$\displaystyle = 3(2x - 1)[x^2 - \frac{2}{3}x + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 - \frac{7}{3}\right]$

$\displaystyle = 3(2x - 1)\left[\left(x-\frac{1}{3}\right)^2-\frac{22}{9}\right]$

$\displaystyle = 3(2x-1)\left(x-\frac{1}{3}+\frac{\sqrt{22}}{3}\right)\left(x-\frac{1}{3}-\frac{\sqrt{22}}{3}\right)$.

4. Touché. (Of course I meant factorable over the rationals. )