 Solution

A cubic over a field is reducible exactly when it has a root in that field: a nontrivial factorization must contain a degree-one factor. By the rational root theorem, a rational root of this monic integer polynomial must be an integer dividing its constant coefficient, hence either $1$ or $-1$. Substitution gives $-2$ and $4$, respectively. Neither is zero. Thus the cubic is irreducible over $\mathbb{Q}$.