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Math Help - HELP...Polynomial Rings Problem...

  1. #1
    Junior Member ginafara's Avatar
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    HELP...Polynomial Rings Problem...

    I am reviewing some problems from polynomial rings and hope that someone can tell me if I am on the right track..

    PROBLEM: Let f(x) \in R[x]. If f(a) = 0, f'(a)=0 show that  (x-a)^{2} divides  f(x) .


    What I have done:

     f(x) = g(x)(x-a)^{2}+r(x) g(x), r(x) \in R[x], deg r(x) < deg (x-a)^{2} or  deg r(x) = 0

    f'(x) = 2g(x)(x-a) + g'(x)(x-a)^{2} + r'(x)

    f(a) = g(a)(0) + r(a) \implies f(a) = r(a) = 0

    so f(x) = g(x)(x-a)^{2}

    f'(a) = 2g(a)(0) + g'(a)(0) +r'(a) \implies f'(a) = r'(a) = 0

    Thus (x-a)^{2} is a factor of f(x) \implies (x-a)^{2}|f(x)

    Is this even close???? Thanks in advance
    Last edited by ginafara; May 28th 2008 at 10:22 PM.
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    In my opinion (it's not the field I prefer), there is a problem :

    Quote Originally Posted by ginafara View Post

    f(a) = g(a)(0) + r(a) \implies f(a) = r(a) = 0

    so f(x) = g(x)(x-a)^{2}
    here, because from r(a)=0, you state that r(x)=0 ,\ \forall x

    Plus, if you go this way, you can directly say that r'(x)=0 since it's the null function ~
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  3. #3
    MHF Contributor kalagota's Avatar
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    i think you are actually done but you missed some reasons.

    note that deg \, r(x) < deg \, (x-a)^2 or deg \, r(x) = 0, that is deg \, r(x) = 0 \, or \, 1

    but the case deg \, r(x) = 1 cannot happen otherwise r'(x) is a constant and that f'(a) \neq 0.

    thus deg \, r(x) = 0, that is r(x) is a constant function. but you have shown that r(a) = 0 and therefore r(x) = 0, \, \forall \, x.
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  4. #4
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    If f(a) = 0 then f(x) = (x-a)g(x).
    Thus, f'(x) = g(x) + (x-a)g'(x) since f'(a) = 0,it follows g(a) = 0 so g(x) = (x-a)h(x).
    And thus f(x) = (x-a)^2h(x).
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