Rings where we cannot factor a polynomial So my question is this. Consider $R[x]$, and say we have an element $\alpha(x)$ of $R[x]$. I know that if we can do the Euclidean algorithm in $R[x]$ then for $c$ a unit with $\alpha(c)=0$ would imply that $(x-c)$ is a factor of $\alpha(x)$ (meaning $\alpha(x)=(x-c)\beta(x)$ for some $\beta(x)\in R[x]$). 
I was wondering: is there a ring $R$ so that $c$ is a unit in $R$, and $\alpha(c)=0$ but we cannot write $(x-c)\beta(x)=\alpha(x)$ for any $\beta(x)\in R[x]$?
 A: The Euclidean algorithm is not needed here nor is it of any relevance whether $c$ is a unit or not. The theorem that if $c \in R$ is a root of $\alpha \in R[x]$, then $\alpha (x)=(x-c)\beta(x)$ for some $\beta(x)\in R[x]$ only depends on division with remainder, which holds in any polynomial ring with coefficients in a ring.
For the proof: Use division with remainder in $R[x]$ to write $\alpha(x)=(x-c)q(x)+r(x)$ where $r(x)$ is the remainder. Evaluate at $x=c$ to obtain $0=r(x)$, but $r(x)$ is at most linear, thus $r = 0$, and the claim follows. 
A: Let $r$ be any element of any commutative ring $R,$ and suppose $f(x)=\sum_{i=1}^n a_i x^n \in R[x]$ is such that $f(r)=0.$ 
\begin{align*}
f(x) &= \sum_{i=1}^n a_i x^i \\
        &=\sum_{i=1}^n a_i (x-r+r)^i \\
 &=\sum_{i=1}^n a_i \sum_{k=0}^i \binom{i}{k} (x-r)^k r^{i-k}\\
        &=(x-r)p(x) + \sum_{i=1}^n a_i r^i\\
        &=(x-r)p(x) + f(r)\\\
        &=(x-r)p(x)
\end{align*}
where $p(x) = \displaystyle \sum_{i=1}^n a_i \sum_{k=1}^i \binom{i}{k} (x-r)^k r^{i-k}.$
The binomial theorem is valid is any commutative ring and and only the first term in the binomial expansions is without a factor of $(x-r).$
