Zero divisor in R[x] for a commutative ring R

Zero divisor in R[x] for a commutative ring R

Given a commutative ring $R$ and a polynomial $f(x)=a_0+\cdots+a_nx^n, \
a_n\neq 0_R$ which is a zero divisor in $R[x]$, I am supposed to show that
$a_n$ is a zero divisor in $R$. Now there exists a polynomial
$g(x)=b_0+\cdots+b_mx^m, \ b_m\neq 0_R$ such that $f(x)g(x)=0_R$ or
$g(x)f(x)=0_R$. In fact, since $R$ is commutative, so is $R[x]$, and so
these two are equivalent. Suppose however, they were not. Then in the
first case, by considering the coefficient of $x^{n+m}$, I would conclude
that $a_nb_m=0_R$. In the second case, I would similarly get $b_ma_n=0_R$.
Either way, $b_m$ is a zero divisor of $a_n$. So why do I need
commutativity?
Thanks!