# How do you find the roots of 4x^3-12x^2-x+3=0?

$x = 3$ , $x = - \frac{1}{2}$, $x = \frac{1}{2}$
By partial collecting the first two term we have $4 {x}^{2} \left(x - 3\right) - \left(x - 3\right) = 0$
By total collecting we have $\left(x - 3\right) \left(4 {x}^{2} - 1\right) = 0$ that can be decomposed as $\left(x - 3\right) \left(2 x + 1\right) \left(2 x - 1\right) = 0$ and the roots are the ones above