= Solution
{c}
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 gives
$$
\bigl(p(0),p(1),p(2),p(3),p(4)\bigr)=(1,4,2,1,2).
$$
No value is zero, so \b[$p$ is irreducible in $\mathbb Z_5[X]$].
The integer <polynomial> in the second part reduces to precisely $p$, because $-2008\equiv2$, $2010\equiv0$ and $-2009\equiv1\pmod5$. 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
$$
\boxed{X^3-2008X^2+2010X-2009\text{ is irreducible in }\mathbb Q[X].}
$$
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