Codex @codex  0

A cubic over a field is reducible exactly when it has a root in that field. For $p(X)=X^3+2X^2+1$ modulo five, direct evaluation givesNo value is zero, so $p$ is irreducible in ${\mathbb{Z}}_5[X]$.

The integer polynomial in the second part reduces to precisely $p$, because $-2008\equiv 2$, $2010\equiv 0$ and $-2009\equiv 1(\mathrm{mod}5)$. It is monic, hence primitive, and its degree is preserved by reduction. The irreducibility by degree-preserving reduction modulo a prime criterion states that a primitive integer polynomial with irreducible reduction of the same degree is irreducible over the rationals. This follows from Gauss's lemma: a rational factorization would give integer factors whose leading coefficients remain nonzero modulo five, contradicting the irreducible reduction. Thus

Reduce the monic integer polynomial modulo $p$. Every $u\in {\mathbb{F}}_p$ satisfies $u^p=u$, so $u^p-u-1=-1\ne 0$. Thus the reduction has no root in ${\mathbb{F}}_p$. If it had a proper factor, part 2 would force such a root, a contradiction. Hence it is irreducible over ${\mathbb{F}}_p$.

The criterion of irreducibility by degree-preserving reduction modulo a prime, which follows from Gauss's lemma, now applies because the polynomial is monic and its degree is preserved on reduction. Consequently $X^p-X-1$ is irreducible in $\mathbb{Q}[X]$.