Understanding a sequence of isomorphisms showing $(2, x)$ is maximal in $\mathbb{Z}[x]$ Aluffi (paragraph III.4.3) gives the following sequence of isomorphisms showing the claim in the title (quoting almost as is):
$$
\frac{\mathbb{Z}[x]}{(2, x)} \cong \frac{\mathbb{Z}[x]/(x)}{(2)} \cong \frac{\mathbb{Z}}{(2)} = \mathbb{Z} / 2 \mathbb{Z},
$$
where $(a_1, \dots, a_n)$ is the ideal generated by $a_1, \dots, a_n$.
I'm trying to trace each step carefully and more or less rigorously, and I'm having trouble somewhere between the first and second step (which is probably more of a notational issue, but I just want to make sure I get the example right).
So I know that $\frac{\mathbb{Z}[x]}{(2, x)} \cong \frac{\mathbb{Z}[x]/(x)}{(2, x) / (x)}$, and I can show that $(\overline{2}) = (2, x) / (x)$, where $\overline 2$ is the class of $2$ in $\mathbb{Z}[x]/(x)$. Thus $(\overline 2)$ is an ideal in $\mathbb{Z}[x]/(x)$, and $(\overline 2) = \{ 2n + (x) \mid n \in \mathbb{Z} \}$.
So, first question: should it really be $\frac{\mathbb{Z}[x]/(x)}{(\overline 2)}$ (NB the overline) on the right-hand side of the first $\cong$?
Assuming it should, and assuming $\mathbb{Z}[x]/(x)$ is factored by $(\overline 2)$, I'm not sure how to justify second isomoprhism. Of course, $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ is obvious, and I could probably prove that $\varphi : (\overline 2) \rightarrow (2); \varphi : 2n + (x) \mapsto 2n$ is an isomorphism (even though I don't really know what an isomorphism of ideals is), but that's my second question: why applying them both at once is justified (assuming that indeed $(\overline 2) \cong (2)$)? Or, otherwise, why is the second $\cong$ justified at all?
 A: Great question. The answer to your first question is yes: the ideal should really be denoted $(\overline{2})$. (This is a mostly harmless bit of sloppiness that many authors will practice.) 
As for your second question, the key is an application of the first isomorphism theorem: 
Claim: Suppose $\varphi \colon R \to S$ is an isomorphism of commutative rings. If $J$ is an ideal of $S$, and $I = \varphi^{-1}(J)$, then $\varphi$ descends to a ring isomorphism $\overline{\varphi} \colon R/I \to S/J$ which sends $x+I$ to $\varphi(x)+J$. 
Proof: Let $\pi_{J} \colon S \to S/J$ be the canonical projection homomorphism. Then the composition $\pi_{J} \circ \varphi  \colon R \to S/J$ is a surjective map with kernel $\varphi^{-1}(\pi_{J}^{-1}(\{0\})) = \varphi^{-1}(J) = I$, so the result follows from the first isomorphism theorem. 
How does this help our situation? Well, Aluffi writes $\mathbb{Z}[X]/\langle X \rangle \cong \mathbb{Z}$, but he leaves the isomorphism implicit. Explicitly, it is induced by the surjective ring homomorphism $e \colon \mathbb{Z}[X] \to \mathbb{Z}$ which sends $f(X) \in \mathbb{Z}[X]$ to $f(0)$. The kernel of $e$ is precisely $\langle X \rangle$, and so the induced isomorphism $\overline{e} \colon \mathbb{Z}[X]/\langle X \rangle \to \mathbb{Z}$ sends the coset $f(X) + \langle X \rangle$ to $f(0)$. To apply our claim above, put $J = 2\mathbb{Z}$, and $I = \overline{\langle 2, X \rangle} \subset \mathbb{Z}[X]/\langle X \rangle$. If $\overline{e}^{-1}(J) = I$, then we're done. I leave this to you! 
