How to factorize the polynomial $ 6x^4 - 11x^3 - 30x^2 + 29x - 6 $

Question

Using an online service I can see that the polynomial $$ 6x^4 - 11x^3 - 30x^2 + 29x - 6 $$ can be factorized as $$ (3 x-1) (2 x-1) (x+2) (x-3) $$ using the Rational Root Theorem, but how is it applied?

Answer 1

Factor both the leading coefficient and the constant at the end. All rational solutions will be of the form $\pm\frac p q$, where $q$ is a factor of the leading coefficient and $p$ is a factor of the constant. From there, guess any of those roots until you get one rights.

Answer 2

From Wiki, if $\frac{p}{q}$ is a factor of $P(x)$, then $p$ is a factor of, in your case, $-6$, and $q$ is a factor of $6$. So, the possibilities are:

$$ 1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, \frac{1}{6}, -\frac{1}{6} \\ 2, -2, \frac{2}{2}, -\frac{2}{2}, \frac{2}{3}, -\frac{2}{3}, \frac{2}{6}, -\frac{2}{6}, \\ 3, -3, \frac{3}{2}, -\frac{3}{2}, \frac{3}{3}, -\frac{3}{3}, \frac{3}{6}, -\frac{3}{6}, \\ 6, -6, \frac{6}{2}, -\frac{6}{2}, \frac{6}{3}, -\frac{6}{3}, \frac{6}{6}, -\frac{6}{6} $$

Now it is just a matter of checking each one of them, it is not that hard because many of the numbers that appear in the list are not unique.