Division Algorithm $\rm\:\Rightarrow\ p(x)\, =\, q(x)\ (x\!-\!c) + r,\ r\in F.\, $ Evaluate at $\rm\ x = c\ \Rightarrow\ r = p(r).\ $ QED
Therefore $\rm\ x - c\ |\ p(x)-p(c)\ $ in $\rm\:R[x],\,$ $\rm \ \forall\ p\in R[x]\:,\ \forall\ rings\ R\quad $ [Factor Theorem]
Said equivalently $\rm\ p(x)\:\equiv\:p(c)\ \pmod{x - c}$
or, in other words, $\rm\ x\,\equiv\,c\ \Rightarrow\ p(x)\:\equiv\: p(c)\,\ $ [also follows by Polynomial Congruence Rule]
E.g. casting nines: mod $\rm 9\!:\ 10\equiv 1\ \Rightarrow\ N= p(10)\equiv p(1) \equiv $ sum of digits of $\rm\, N\,$ in radix $10$.
Note how the result is much clearer in the language of congruence arithmetic.
Remark $\ $ The shift automorphism $\rm\ x\to x+c\ $ reduces the proof of the Factor Theorem to the "obvious" special case $\rm\:c = 0,\:$ e.g. see here and my sci.math post appended below. However, this approach is a bit risky pedagogically since such a proof is not completely rigorous without knowledge that such maps are ring automorphisms. It is essential that students learn how to make rigorous (ring-theoretically!) prior informal arguments about substitution, changing "variables", etc. Much subtlety lies here, e.g. even in the informal notation for polynomials, such as $\rm\: P(X\!+\!c)\:$ below. This automorphism is the essence behind Patrick's answer, and is also implicitly in Pierre's. Of course such shifting is yet another example of transformation-based problem-solving, cf. my recent post on analogously applying a shift so that Eisenstein's irreducibility criterion applies.
asdf wrote to sci.math on 29 Mar 2006 (paraphrased)
How do you prove that for a polynomial P(X)
$\rm P(c)=0\ \Rightarrow\ X-c\ |\ P(X)\ $ i.e. $\rm\ (X-c)\ Q(X)\ =\ P(X)\ $ for some $\rm\ Q(X)\ $
For $\rm\ c=0\ $ it specializes to the obvious case: $\rm\ X\mid P(X) \iff P(0)=0$
If $\rm\ c\ne 0\ $ reduce to $\rm\ c=0\ $ by a shift: $\rm\ X\!-\!c\mid P(X) \iff X\mid P(X\!+\!c) \iff P(c)=0 $