 Solution

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