If $R$ is a ring, and $a$ in $R$ is a root of $F$ in $R[X]$ then is it true that there exists $G$ in $R[X]$ such that$ F= (X-a)G$?

Question

If $R$ is a ring, and $a$ belongs to $R$ such that $a$ is a root of $F$ in $R[X]$ then is it true that there exists $G$ in $R[X]$ such that $F= (X-a)G$ ?

Answer 1

Write $F(X) = b_0 + b_1 X + \dots + b_n X^n$, let $a \in R$ be a root of $F$, then $F(a) = 0$, therefore $$F(X) = F(X) - F(a) = b_1 (X-a) + b_2 (X^2 - a^2) + \dots + b_n (X^n - a^n)$$

And for any $k$, $$X^k - a^k = (X-a) (X^{k-1} + a X^{k-2} + \dots + a^{k-1})$$

so if you factor the first equation you get $F(X) = (X-a) G(X)$ for some $G$.

Answer 2

Hint $\ {\rm mod}\ x\!-\!a\!:\,\ \color{#c00}{x\equiv a}\ \Rightarrow\ F(\color{#c00}x)\equiv F(\color{#c00}a)\equiv 0$