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Math Help - Ring of Polynomials

  1. #1
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    Ring of Polynomials

    Assume that (x-a) | g(x) in the ring of polynomials R[x]. Prove that (x-a)^2 | g(x) if and only if (x-a) |
    g' in R[x].
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  2. #2
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    A start

    If (x-a)^2 | g(x) then g(x) = (x-a)^2 p(x) for some polynomial p(x). Take the derivative of g(x) and see if (x-a) | g'(x)
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  3. #3
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    Okay, I can see that, but I am still stuck on the "if and only if" part. If (x-a) | g', then g' = (x-a)q(x) for some polynomial q, but then how exactly would I integrate that? Would it just be g = ((x-a)^2)q(x)/2 + b, and if so, how would I deal with the "+b"? or do I have to treat the q(x) as more than a constant like I suspect, and if so, how?
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  4. #4
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    Hi

    You assumed (x-a)|g(x), i.e. there is a p(x) such that g(x)=(x-a)p(x).

    We have g'(x)=(x-a)p'(x)+p(x) hence p(x)=g'(x)-(x-a)p'(x).

    Therefore: (x-a)|g'(x)\Rightarrow (x-a)|p(x)\Rightarrow (x-a)^2|g(x)
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