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Thread: A doubt on divisibility

  1. #1
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    A doubt on divisibility



    Why is it so? (Check out the image first)


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    Re: A doubt on divisibility

    If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?
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    Re: A doubt on divisibility

    Claim:
    $$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = x^{2n+1}+a^{2n+1}$$

    Proof:
    $$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = (x+a)(x^{2n}-ax^{2n-1}+a^2x^{2n-2}\pm \cdots - a^{2n-1}x+a^{2n})$$

    Now, let's expand and line up terms (the top line is the summation multiplied by $x$ while the bottom line is the summation multiplied by $a$):

    $$\begin{matrix} & x^{2n+1} & -ax^{2n} & + a^2x^{2n-1} & \pm \cdots & -a^{2n-1}x^2 & +a^{2n}x & \\ + & & ax^{2n} & -a^2x^{2n-1} & \pm \cdots & +a^{2n-1}x^2 & -a^{2n}x & +a^{2n+1}\end{matrix}$$

    Note that the central terms all cancel out, and all that is left is $x^{2n+1}+a^{2n+1}$.

    Can you do something similar for $x^{2n}-a^{2n}$?
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    Re: A doubt on divisibility

    Quote Originally Posted by MarkFL View Post
    If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?
    Oh yeah got it. To make f(-a) zero,we need to ... you know
    Got it got it


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    Re: A doubt on divisibility

    Hey i have 17 questions as doubt from my text book.can i send it all to this thread



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    Re: A doubt on divisibility

    Quote Originally Posted by sbjsavio View Post
    Hey i have 17 questions as doubt from my text book.can i send it all to this thread



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    We would prefer separate threads for each question.
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    Re: A doubt on divisibility

    Quote Originally Posted by SlipEternal View Post
    We would prefer separate threads for each question.
    Ok then



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    Re: A doubt on divisibility

    Quote Originally Posted by sbjsavio View Post
    Ok then



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    And post work on each question so we may see where you are struggling. Don't post a problem without any work/attempts shown.
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