Zero divisor in $R[x]$ Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$?

This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.

A ring $R$ is called Armendariz if whenever $f(x)=\sum_{i=0}^{m}a_ix^i, g(x)=\sum_{j=0}^{n}b_jx^j \in R[x]$ such that $f(x)g(x)=0$,  then $a_ib_j=0\ \forall\ i,j$.

In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result. 
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\\ a_1b_0+a_0b_1=0 ,\\ \vdots \\a_nb_0+\dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0\implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 \implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 \implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 \implies a_1b_0=0$.
Similarly we get $a_ib_0=0\ \forall\ 1\leq i \leq n$.  
Now original equations reduces to
$$a_0b_1=0\\ a_1b_1+a_0b_2=0\\ \vdots\\ a_{n-1}b_1+\dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0\ \forall\ 1\le i \le n$.
Similarly, Repeating it we get $a_ib_j=0\ \forall\ 1 \leq i,j \leq n$. $\hspace{5.5cm}\blacksquare$
Some other examples of Armendariz Rings are:


*

*$\Bbb{Z}/n\Bbb{Z}\ \forall\ n$.

*All domains and fields (which are reduced).

*If A and B are Armendariz , then $A(+)B$ in which multiplication is  defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.

*Direct Product of Armendariz rings.

