$\left<2,x\right>$ is a maximal ideal of $\Bbb Z[x]$ 
I want to show that $\left<2,x\right>$ is a maximal ideal of $\Bbb Z[x]$. 

My game plan is to use the 3rd isomorphism theorem to somehow get that $Z[x]/\left<2,x\right>$ isomorphic to $Z_2$ (since every this would mean that $Z[x]/\left<2,x\right>$ is a field and hence $\left<2,x\right>$ would be a maximal ideal) but I'm not entirely sure how to arrive at that conclusion. 
My first thoughts are that since $\left<x\right>$ is an ideal of $Z[x]$ and $\left<x\right>\subset\left<2,x\right>$ we have that $Z[x]/\left<2,x\right> \cong Z[x]/\left<x\right>/\left<2,x\right>/\left<x\right>$ by the 3rd isomorphism theorem, but I'm not sure if this is the right direction. Any hints or tips would be great. 
 A: This seems like a possible direction. As a next step, you can say that:
$$ \mathbb{Z}[x]/(2,x) \simeq \left(  \mathbb{Z}[x]/(x)\right) /\left((2,x)/(x) \right) \simeq  \mathbb{Z}/(2) = \mathbb{Z}_2,$$
and you're done. The second isomorphism needs some justification, perhaps. This would involve either waiving your hands around it, or doing it by hand.
You can also refer to the second isomorphism theorem. Because $ \mathbb{Z}[x] =  \mathbb{Z} + (2,x)$, we have that:
$$   \mathbb{Z}[x]/(2,x) = \left(  \mathbb{Z} + (2,x) \right) /(2,x) \simeq  \mathbb{Z}/\left(  \mathbb{Z} \cap (2,x) \right) \simeq  \mathbb{Z}/(2) = \mathbb{Z}_2.$$
A: You can look at something constructive, if you may. Note that $P\in\Bbb Z[X]$ is in $(2,X)$ if and only if its constant coefficient is divisible by $2$, that is, the elements of $(2,X)$ are of the form $2a_0+XP$ with $a_0\in\Bbb Z$ and $P\in\Bbb Z[X]$. Then consider the surjective homomorphism $\Bbb Z[X]\to \Bbb Z/(2)$ given by $P\mapsto { P(0)}+(2)$, and recall $\Bbb Z/(2)$ is a field. 
In general, we have that $$\frac{R[x]}{(I,x)}\simeq \frac{R}{I}$$ 
