All ideals of $\mathbb Z[x]$ are of the form $\langle n\rangle\langle f(x)\rangle, \langle n,f(x)\rangle$, attempt inside but not complete How can we show this is true? Here $f(x)$ is any degree $1$ and higher polynomial in $\mathbb Z[x].$ I've tried saying suppose I is not equal to $\langle n\rangle$ or $\langle f(x)\rangle$, and then through a series of arguments show that it must be $\langle f(x),n\rangle$ for some f and n.I have chosen f to be the smallest positive degree polynomial in my ideal. Then choose an element in the ideal that is not a multiple of $f(x).$ Then $f$ doesn't divide $g,$ and hence $g= fq + r$, where $r$ must be of degree 0 by minimality of the degree of $f.$ So then we have a constant $r$ (not equal to $0$ or $1$) that's in the ideal. Now any other poly $h$ in I that isn't a multiple of $f$ must have remainder a constant by the same reasoning. Now if this constant remainder is coprime with $r,$ then $1$ is in the ideal and we get a contradiction. So only elements whose constant remainder with $f$ has a factor in common with $r$ are in $I$. If $r$ is prime, we would be done as then all remainders would be a multiple of $r$, hence we would have
$I= \langle f(x), r\rangle$. But if $r$ is not prime, I don't know how to proceed. Could someone help or give an easier proof? Thanks! 
Edit: Sorry there was formatting issues, I've fixed it.
 A: Hmm, you have a problem with your use of division.  For the division-with-remainder to work in $\Bbb{Z}[x]$ you need the denominator $f$ to be monic (leading coefficient 1, or at least leading coefficient a unit), but it might not be.  Can you write $(4,2x,x^2)$ in your desired format?  So I think you are trying to prove something that isn't true.
A: To complete your proof, think about the collection of all the remainders you get from all polynomials which are not multiples of $f$; they form an ideal in $\mathbb{Z}$, and all ideals in $\mathbb{Z}$ are generated by one integer (the greatest common divisor). So you can still get an integer $n$.
You should be a bit more careful with your statement $g = fq+r$ though; consider the case $f = 2x$ and $g = 3x^2 + 1$. How can you express $g = fq+r$?
If you know about Gauss's Lemma, you can modify your proof to consider polynomials in $\mathbb{Q}[x]$, where you can divide polynomials like this, and then argue from them to polynomials in $\mathbb{Z}[x]$.
