$f$ has root $\alpha$, then $f = (X-\alpha)g$ for some $g$ I need some help with the following problem:
Suppose $R$ is a unique factorisation domain and $f \in R[X]$ such that $\deg f > 0$ and $f$ has a root $\alpha \in R$. Then $f = (X-\alpha)g$ for some $g \in R[X]$. 
My idea so far was to write $f = \sum_{j = 0}^n a_jX^j$ and $g = \sum_{j = 0}^{n-1} b_jX^j$. If I compute $(X-\alpha)g$ I obtain
\begin{align*}
-\alpha b_0 + (b_0-\alpha b_1)X+(b_1-\alpha b_2)X^2+\ldots+(b_{n-2}-\alpha b_{n-1})X^{n-1}+b_{n-1}X^n.
\end{align*}
When I compare the coefficients I get $a_0 = -\alpha b_0$ etc.
But if I want to show that $g \in R[X]$ I need to show that $b_i \in R$ for all $i$, and to reach that I need to know if $\alpha^{-1}$ exists in $R$. 
Maybe the whole approach is wrong.
Any help is appreciated, thanks.
 A: Hint: In $R[X]$ we have long division of polynomials. For $a,b\in R[X]$ with $b$ having a unit as leading coefficient we have $a=qb+r$ for some $q,r\in R[X]$ with $\deg r < \deg b$.

The important ingredient is $b$ having a unit as leading coefficient:
Let $a=\sum_{i=0}^n a_i X^i$ with leading coefficient $a_n\neq0$, $b=\sum_{i=0}^m b_i X^i$ with leading coefficient $b_m\in R^*$ a unit.
If $n<m$ we have $a=qb+r$ for $q=0$, $r=a$ with $\deg r = \deg a = n < m = \deg b$.
If $n\ge m$ we can cancel the leading coefficient by substracting $a_n b_m^{-1} X^{n-m}r$, so we have
$$
\deg\big( a - a_n b_m^{-1} X^{n-m}r\big) < n.
$$
Continue this way to reduce the degree below $m$ by substracting multiples of $r$.
All you need is a commutative ring with unit and the leading coefficient of $b$ to be a unit.
A: Suppose $f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x+a_0.$ Then $$f(x)-f(a) = a_n (x^n-a^n) + a_{n-1}(x^{n-1}-a^{n-1}) + \cdots + a_1(x-a).$$ Now since $x^k-a^k = (x-a)(x^{k-1} + x^{k-2}a + \cdots + a^{k-1})$ we can factor out $(x-a)$ from every term on the right hand side, so we have $$f(x) = (x-a)g + f(a).$$
Note, we don't need $R$ to be a UFD or even a domain - all we needed was that $R$ is a commutative ring.
