Procedure for showing a polynomial is irreducible in $\mathbb{Q[x]}$ Let $f(x) = x^5 − 6x^2 + 21x + 13$
What is the procedure for showing $f(x)$ is irreducible in $\mathbb{Q}[x]$?
 A: Suppose $f(x) = p(x) q(x)$ over $\mathbb{Z}$, where we can take $p$ and $q$ monic and non-constant. Use subscript $l$ to indicate taking coefficients mod $l$. Note that $\deg p_l(x) = \deg p(x)$, etc., and $f_l(x) = p_l(x) q_l(x)$.
One computes
$$ f_2(x) = x^5+x+1 = (x^2+x+1)(x^3+x^2+1) $$
$$ f_3(x) = x^5 + 1 = (x+1)(x^4-x^3+x^2-x+1) $$
The factors of $f_2(x)$ have no roots, so are irreducible. The quartic factor of $f_3(x)$ also has no root, so if it factors, it's the product of monic irreducible quadratics. The quadratics must have discriminant $-1$ since otherwise they would factor with the quadratic equation, hence we get factors of the form $x^2+bx+c$ where $b^2-c = -1$. The possible quadratics are $x^2+1$, $x^2+x-1$, $x^2-x-1$, and no two of these give the quartic as product, so the quartic is irreducible
Hence the degrees of $p$ and $q$ are ($2$ and $3$ from $l=2$) and ($1$ and $4$ from $l=3$), a contradiction, so $f$ is irreducible over $\mathbb{Z}$, hence over $\mathbb{Q}$.
Maybe you can find a less computational way to show the quartic is irreducible....
