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Math Help - [SOLVED] Help for a proof : polynomial divisibility

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Help for a proof : polynomial divisibility

    I have to demonstrate that " P(x) is divisible by (x-\alpha) if and only if p(\alpha)=0."
    My work : If P(x) is divisible by (x-\alpha), then I can write P(x)=(x-\alpha)\cdot R(x).
    If P(\alpha)=0, then (\alpha-\alpha)\cdot R(\alpha)=0, which is true. So \Leftarrow) is proved.
    Now I have to show "if P(x) is divisible by (x-\alpha), then P(\alpha)=0." I don't know how to continue this.
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  2. #2
    Eater of Worlds
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    I am going to use (x-c) instead of alpha. OK?

    Let's use the remainder theorem.

    Then, P(x)=(x-c)q(x)+P(c)

    For some quotient q(x).

    If P(c)=0, the P(x)=(x-c)q(x); that is, x-c is a factor of P(x).

    Also, if x-c is a factor of P(x), then the remainder when we divide P(x) by

    x-c must be 0. Therefore, by the remainder theorem P(c)=0.

    Is that what you were getting at?. Unless you want to be a purist and prove the remainder theorem as well.

    Is that want you wanna do?.
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  3. #3
    MHF Contributor arbolis's Avatar
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    I am going to use (x-c) instead of alpha. OK?
    O.K.
    I just understood your proof. I knew the remainder theorem, but not applied to polynomials! Thanks, I got it...
    Unless you want to be a purist and prove the remainder theorem as well.

    Is that want you wanna do?.
    No! Thank you enough! I've already done that last year (in algebra), but for any number in Z. The proof I had to do now is an exercise I got in Calculus II, so I won't be a purist when it comes to algebra, ahaha.
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