The ideal generated by $2$ and $ x$ in $\mathbb{Z}[x]$ Is the ideal generated by $2$ and $x$ in $\mathbb{Z}[x]$ all polynomials except for the odd constants?
And is this ideal a non-maximal, prime ideal?
 A: No.
The polynomial $x+1$ is not in $(2,x)$. 
A: It is all polynomials with even constant terms. To show this, pick a polynomial $a_n x^n + a_{n-1} x^{n-1} + \cdots +a_1 x + 2a_0$. This is equal to $x\left(a_n x^{n-1} + a_{n-1} x^{n-2} + \cdots +a_1 \right) + 2a_0$, so it's in the ideal.
Showing that polynomials with odd constant terms is just doing this backwards (and shifting the indexing on $a$ appropriately).
EDIT: My ring theory is terrible, but I believe it is maximal.
Say there is an ideal $I$ such that $(2,x) < I$. Then it must contain a polynomial with odd constant term. Let's call it $a_n x^n + a_{n-1} x^{n-1} + \cdots +a_1 x + 2a_0 + 1$. Because $I$ contains $x$ and is contagious under multiplication, $x \left (a_n x^{n-1} + a_{n-1} x^{n-2} + \cdots +a_1 \right) = a_n x^n + a_{n-1} x^{n-1} + \cdots +a_1 x$ is in $I$. $I$ is closed under subtraction, so $2a_0 + 1$ is in $I$. Following the same logic with $2$, we see that $1 \in I$, making $I = \mathbb{Z}[x]$.
Also, maximal ideals are prime in a commutative ring with $1$.
