Let $A$ be a commutative ring and $f\in A[X]$ with $f \not\equiv 0$. Show that if $f|0$ in $A[X]$ then $\exists a \in A$ st $a \neq 0$ and $af = 0$

Question

For example, let $f= \sum_{i=0}^n a_ix^i$. If every $a_i\not\in U(A)$, then $\exists b_i\in A$ with $b_i \neq 0$ for every $i$ such that $a_ib_i=0$, so I can take $a = \prod b_i$ reaches what I need.

I tried using $f|0$ saying that exists a $g$ such that $fg\equiv0$, but the only thing I've reached with this is this is that the independant term of $fg$ is $0$.

Another thing I noticed is that $Im(f) \in A-U(A)$