The rational root test

Candidates. $p\in\{\pm1,\pm2,\pm3,\pm6\}$, $q\in\{1,2\}$, so possible roots: $\pm1,\pm2,\pm3,\pm6,\pm1/2,\pm3/2$.

Test.
- $P(1)=2+1-7-6=-10$.
- $P(-1)=-2+1+7-6=0$. Root.
- $P(2)=16+4-14-6=0$. Root.
- $P(-2)=-16+4+14-6=-4$.
- $P(3)=54+9-21-6=36$.
- $P(1/2)=1/4+1/4-7/2-6=-8$.
- $P(-3/2)=-27/4+9/4+21/2-6=-3$.

Factor. Dividing by $(X+1)(X-2)=X^2-X-2$: $P=(X+1)(X-2)(2X+3)$.

Conclusion. Rational roots: $-1$, $2$, $-3/2$.