How to show that the following ideal is prime/maximal? If $I$ is the set of polynomials that can be written as $2p(x)+(x^2+x+1)q(x)$, how can I show that it is or isn't a prime ideal of $R=\mathbb{Z}[x]$? 
$I$ know that if $I$ consider two polynomials A and B such that $AB\in I$, at least one of them must also be in $I$ if $I$ is prime. Not sure how to actually prove this though.
Following this, how could I show that it is or isn't maximal?
 A: An ideal $I$ is prime iff $R/I$ is an integral domain. An ideal $I$ is maximal iff $R/I$ is a field.
(these observations come in handy in both directions - sometimes it is more natural to show that $I$ is prime (say if it is generated by a prime element), sometimes that $R/I$ is a domain.)
This way you can use some observations about taking quotients, like isomorphism theorems.
For example, in your case where $I=(2, x^2+x+1),$ we have that
$R/I=\mathbb{Z}[x]/(2, x^2+x+1) \simeq \big(\mathbb{Z}[x]/(2)\big)/\big((2, x^2+x+1)/(2)\big) \simeq \mathbb{Z}_2[x]/(x^2+x+1)$
and now, it remains to check that $x^2+x+1$ is irreducible over $\mathbb{Z}_2$, hence prime, hence the quotient ring is an integral domain.
I would suggest that you try to convince yourself that the claimed isomorphisms indeed exist. 
Also, if you are not familiar with these quotient constructions, I apologize, for this answer is then completely useless. 
By the way, you can even see that the quotient is a (finite) field.
A: If you know quotient rings then you should proceed as in Pavel's answer. If not there is a simple direct proof that $\, I = (2,x^2\!+x+1)\,$ is maximal in $\,\Bbb Z[x].\,$  Every $\,f\in\Bbb Z[x]\,$ is congruent mod $\,I\,$ to $\,\bar f =  ax+b,\,$ for $\,a,b\in \{0,1\},\,$ by taking the remainder of $\,f\,$ mod $\,x^2\!+x+1\,$ then  mod $2.\,$ So $\, f\not\in I\,\Rightarrow\, \bar f \equiv 1,\,\color{#c00}x,\,$ or $\,\color{#0a0}{x\!+\!1}.\,$ In all $\,3\,$ cases $\,I+(f) = I+(\bar f ) = 1,\,$ so $\,I\,$ is maximal, since $\,(\color{#c00}x),(\color{#0a0}{x\!+\!1})\,$ are comaximal to $\,(x^2\!+x+1),\,$ by $\,x^2\!+x+1 - \color{#c00}x(\color{#0a0}{x\!+\!1}) = 1$
Remark $\ $ This is essentially a special case of the general theorems mentioned by Pavel. You may find it instructive to compare this special case to the general results to help gain better intuition on the more general abstractions. 
