$p(x)$ 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.

Here's how it goes : the polynomials $\{1, (x-c), (x-c)^2, \dots \}$ form a basis of the vector space $F[x]$. Write
$$
p(x) = a_0 + a_1 (x-c) + a_2 (x-c)^2 + \dots + a_n (x-c)^n.
$$
Then 
$$
p(x) = (x-c) \left( a_1 + a_2(x-c)^2 + \dots + a_n (x-c)^{n-1} \right)+ a_0 
$$
and you can see that $p(c) = a_0$. 
Hope that helps,
