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


    What I have done:

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

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

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

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

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

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

    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

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

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

    Plus, if you go this way, you can directly say that $\displaystyle r'(x)=0$ 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 $\displaystyle deg \, r(x) < deg \, (x-a)^2$ or $\displaystyle deg \, r(x) = 0$, that is $\displaystyle deg \, r(x) = 0 \, or \, 1$

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

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