$f(x)$ has factor $x-a$ iff $R$ is UFD?

Question

Let $R$ be a commutative ring with identity. What is the condition on $R$ so that the following statement is valid?

Statement: Let $f(x) \in R[x]$. Then $f(x)$ has factor $x-a$ if and only if $f(a)=0$ for some $a \in R.$

I think it's possible for $R$ to be a Unique Factorization Domain.

Thanks for looking at it.

score 2 · Accepted Answer · answered Oct 15 '15 at 16:09

It holds for every commutative ring with identity, because for all $R$ and $a \in R$ we have the identity

$$f(x) = q(x) (x-a) + f(a)$$

with the quotient $q(x) \in R[x]$ from the polynomial long division algorithm, or from $f(x) \equiv f(a) \mod (x-a)$.

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