Let $\ g = X^3\ -9X + 16 $. Prove that $g$ is irreducible over the rational numbers.
So far I have used reduction modulo $5$ and this gives $g_5 = X^3 +X + 1$. Then I get $$ g_5(0) \equiv 1 \pmod5,\\ g_5(1) \equiv 3 \pmod5,\\ g_5(2) \equiv 1 \pmod 5,\\ g_5(3) \equiv 1 \pmod 5,\\ g_5(4) \equiv 4 \pmod 5.$$
As none of these are equal to $0\pmod5$ does this mean that $g$ is irreducible?
Also would this work modulo $2$ or $3$? How do I know which prime to pick?