Given a polynomial function, f(x), with degree n>=3. For any real number, a, divide f(x) by (x-a)^(n-1)
This will ensure that the quotient, q(x), is linear and that the remainder, r(x), is of degree n-2.
1) f(a) = r(a) and 2) f '(a) = r '(a) giving the tangent as y - r(a) = r '( a)[x - a]. See if you can prove this - it's pretty easy. This really struck me as f(x) and r(x) are clearly not the same polynomial and yet! f(a) and r(a) yield the same value as well as f '(a) and r '(a).
This all started with Dr. Barker's post: