The easiest way, but unfortunatelly only works some times is the following:
Step 1: Guess a root $a$ of $P(x)$. Step 2: Divide by $x-a$. Step 3: Factor the remaining quadratic.
The following usually helps:
Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. If all coefficients are integers, and $r=\frac{p}{q}$ is a irreducible rational root, then $p|a_0$ and $q|a_3$.
So for your problem, the only possible rational roots are $\pm 1, \pm 2, \pm 3$ and $\pm 6$, either one works or you need to rely on the cubic formula (which you can find easely but it is hard to memorize)...