Let N = {f in Z[x] | f has constant term 0}, clearly N is an ideal of Z[x], show that N is a maximal ideal of Z[x].
I don't beleive it is a maximal Ideal... Here is why
Define a Homomorphism from $\displaystyle \phi:\mathbb{Z}[x] \to \mathbb{Z}$ by $\displaystyle \phi(f) \to f(0)$
Note that the kernel of $\displaystyle \phi$ is N
$\displaystyle \phi $ is onto so we can use the first isomorphism theorem to get
$\displaystyle \mathbb{Z}/N \cong \mathbb{Z}$
Since $\displaystyle \mathbb{Z}$ is an Integral Domain this implies that The Ideal is Prime and not maximal.
This ideal in $\displaystyle \mathbb{Z}[x]$ is contained in the Ideal gentered by (2, and the same set f) and this ideal is maximal