How to analyze the elements in a polynomial quotient ring?

I am looking at some old exams, and one thing is to prove that in a quotient ring $$R = F[x]/\langle p(x)\rangle$$, where $$F$$ is a field and $$\langle p(x)\rangle$$ is the ideal generated by a polynomial $$p(x) \in F[x]$$, that all elements are either units or zero divisors. I have tried looking this up and read several documents on quotient rings, but I haven't seen a mention of this fact.

Here's my attempt at a proof. Suppose $$a(x) + \langle p(x)\rangle$$ is not a unit. Then this element generates a proper ideal in $$R$$. This corresponds to a proper ideal in $$F[x]$$ which contains $$a(x)$$ and $$p(x)$$. $$F[x]$$ is a PID, so this ideal is generated by some other polynomial $$f(x)$$, which cannot be a unit in $$F[x]$$. Honestly, I'm not sure where to proceed, and I'm a little tired and scared preparing for this exam.

Any hints would be appreciated, more than a full solution.

Thanks

Say you take an element $$\overline{q(x)} \in {F(x)/\langle p(x)\rangle}$$. You have two cases:
Case I: $$q(x)$$ does not share any common factor with $$p(x)$$ i.e. they are coprime to each other. In this case, the Euclidean algorithm gives you that there are $$a(x), b(x)$$ such that $$a(x) p(x) + b(x) q(x) = 1$$. This means that
$$b(x) q(x) \equiv 1 ($$ mod $$p(x))$$ which exactly means $$\overline{q(x)}$$ is a unit.
Case II: $$q(x)$$ shares a factor $$c(x)$$ with $$p(x)$$ i.e. $$q(x) = c(x) d(x)$$ and $$p(x) = c(x) e(x)$$ for some $$d(x), e(x)$$.
In this case, you see that $$q(x) e(x) \equiv 0 ($$ mod $$p(x) )$$ which means that $$\overline{q(x)}$$ is a zero-divisor.