Two standard exercises in Hungerford, 3rd ed.
Ex 4.1.13 Let R be a commutative ring. If $a_n \neq 0$ and $f(x)=a_0+a_1x+...+a_n x^n$ is a zero divisor in $R[x]$, prove that $a_n$ is a zero divisor in $R$.
Ex 4.1.15 (a) Let $R$ be a commutative ring with identity and $a \in R$. If $a^3=0$, show that $1+ax$ is a unit in $R[x]$.
Ex 4.1.15 (b) Let $R$ be a commutative ring with identity and $a \in R$. If $a^4=0$, show that $1+ax$ is a unit in $R[x]$.
Why does Hungerford insist on $R$ being commutative in the three cases above? According to his definitions:
An element $a$ in a ring $R$ with identity is called a unit if there is $u \in R$ such that $au=1=ua$, and
An element a in a ring $R$ is a zero divisor provided that $a \neq 0$ and there is $c \neq 0$ such that $ac=0$ or $ca=0$.
In exercise 4.1.13, there is $g(x) \in R[x]$ such that $f(x)g(x)=0$ or $g(x)f(x)=0$, and following multiplication we have $a_n b_m =0$ or $b_m a_n = 0$ some $0 \neq b_m \in R$.
In exercise 4.1.15 (a), $g(x)=1-ax+a^2x^2$ is an inverse$
In exercise 4.1.15 (b), $g(x)=1-ax+a^2x^2-a^3x^3$ is an inverse.
Have I made a calculation error, misread the definitions or does Hungerford impose something stronger to spare the reader some time in multiplying twice?