Thread: prove a simple property of factors of polynomials

Originally Posted by eniuqvw

Let f(x) be a polynomial of degree n. c(sub 0)x^0 + ... + c(sub n)x^n


I need to prove that if n >= 1 and f(a) = 0 from some real a, then f(x) = (x - a)h(x) where h is a polynomial of degree n - 1. The book i'm using (calculus vol 1 by Apostol) gives the hint: "consider p(x) = f(x + a)" - but I still can't come up with a proof.

Couldn't you just use the fundamental theorem of Polynomials?

That for every polynomial of degree there are exactly complex roots, some of which may be repeated.

well, I'm working under the assumption that we're supposed to use only what we've been given to work with. somehow i need to show that f(x)/(x-a) is a polynomial of degree n - 1. I wonder if there's a way to show this (somehow using the fact that f(x + a) is an nth degree polynomial.) without using the fundamental theorem or division algorithm.

I just found a post in some physics forum in 2006 where they're trying to prove the same thing. They showed that x | f(x + a) implies that (x - a) | f(x) (obviously f((x - a) + a) = f(x)). Expanding the polynomial f(x + a) all the constant terms go away because, by assumption, f(a) = 0, so it's true that x divides f(x + a). By modus ponens, (x - a) | f(x). But I'm not sure how to show that f(x)/(x - a) is a polynomial of degree n - 1? How does this follow from the fact that x - a divides f(x) (where f(x) is a polynomial of degree n)?

the hint tell you that p(0)=0, thus p is a polynomial of degree n but without constant term.
thus f(x+a)=p(x)=xg(x) for some g of degree n-1.
therefore f(x)=(x-a)g(x-a).