$p(x)$ divided by $x-c$ has remainder $p(c)$? [Polynomial Remainder Theorem] This is from Pinter, A Book of Abstract Algebra, p.265. 
Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.
This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.

By the division algorithm, if $a(x)$ and $b(x)$ are any polynomials, and $a(x)\neq 0$, then there exist unique $q(x)$ and $r(x)$ such that
$$b(x) = q(x)a(x) + r(x),\qquad r(x)=0\text{ or }\deg(r)\lt \deg(a).$$
Let $b(x) = p(x)$, and $a(x)=x-c$. Then $r(x)$ must be constant (since it is either zero or of degree strictly smaller than one), so
$$b(x) = q(x)(x-c) + r.$$
Now evaluate at $x=c$.
Note. I find it strange that you say that this is "easy if $c$ is a root of $p(x)$". The Factor Theorem (that $x-c$ divides $p(x)$ when $c$ is a root of $p(x)$) is a corollary of this result. How exactly do you prove it without this?
