I'm asked to show that $27X^3 - 13X^2 + 180 \in \mathbb{Q}[x]$ is irreducible in $\mathbb{F}_{13}[X], \mathbb{Z}[X]$ and $\mathbb{Q}[X]$. I've managed to proof the first two. In $\mathbb{F}_{13}[X]$ I checked that it has no roots and therefore is irreducible. This also shows that is irreducible in $\mathbb{Z}[X]$. Normally I would now apply Gauss's lemma, which tells me that it is in fact irreducible in $\mathbb{Q}[X]$. The problem is, my text book says I can only use this result if the polynomial is monic, which is not the case here. Altough Wikipedia says a polynomial doesn't have to be monic, I'm only allowed to use my text book as reference.
I tried a general factorization: $$ 27X^3 - 13X^2 + 180 = (27X-a)\cdot (X^2 + bX + c), $$ but if I work this out, I get the original equation. Another way is to check every rational root, using the fact that if $\frac{b}{c}$ is a root (with $b$ and $c$ coprime), then $b$ divides $180$ and $c$ divides $27$. But these are just too many options to check by hand.
Is there any way I can show it is irreducible in $\mathbb{Q}[X]$ without the need to proof a generalisation of Gauss's lemma, or is that the only possibility?
Hint $\ $ If it were reducible over $\Bbb Q$ it would have a rational root, which, by the Rational Root Test, has denominator $3^n$ dividing $27$, which would yield a root mod $13$, since $3^n$ is invertible mod $13$.