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 \b[the cubic is irreducible over $\mathbb Q$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 2F / ii / Solution

 Solution (source code)