I have a somewhat silly sounding question:
Let $R$ be an arbitrary commutative ring with $1$. Let $f \in R[X]$.
a) $f(1) = 0 \Rightarrow \exists g \in R[X]: f= (X-1)g$
b) for some $a \in R$: $f(a)= 0 \Rightarrow \exists g \in R[X]: f= (X-a)g$
It's not difficult to show that it is $f =(h + X - 1)g$ for a nilpotent $h \in R[X]$. From this it also follows quickly that $cf= (X-1)(cg)$ for some $c \in R$. That's everything I can show so far. Question might sound silly, but I'm thankful for any input.
Remark: although the division algorithm works in all rings, decomposition in arbitrary rings will not be unique.