Thread: prove a simple property of factors of polynomials

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.

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 $\displaystyle n$ there are exactly $\displaystyle n$ 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).