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Math Help - prove a simple property of factors of polynomials

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by eniuqvw View Post
    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 n there are exactly n complex roots, some of which may be repeated.
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  3. #3
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    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.
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  4. #4
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    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)?
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  5. #5
    Senior Member Shanks's Avatar
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    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).
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