[emerald000]

Originally Posted by SKG_Scintill

Double post, but is "b" always f(a)?

I totally forgot about that theorem. Here's a proof:

Let f be a polynomial of degree n. The formula for dividing f(x) by (x-a) is:

f(x) = g(x)(x-a) + b, where g(x) is a polynomial of degree n-1 and b is the remainder of the division.

Let's see what is the value of f(a):

f(a) = g(a)(a-a) + b
f(a) = g(a)(0) + b
f(a) = b

QED