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Math Help - Really Easy Question about Zeroes of a Polynomial

  1. #1
    Senior Member slevvio's Avatar
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    Really Easy Question about Zeroes of a Polynomial

    Let \mathbb{F} be a field, let f \in \mathbb{F}[x] be a polynomial, and let \alpha \in \mathbb{F}.

    Then I understand that if (x-\alpha) | f, then f(\alpha) = 0, this is obvious.

    But it doesn't seem obvious to me that if f(\alpha)=0, then (x-\alpha)|f. How can I show this? Is there some result about factorisation that implies this?

    Any help would be appreciated.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Since

    \mathbb{F}[x]

    is an integral domain we can use the division algorithm.

    Now lets divide

    f(x) \text{ by }  (x-\alpha)

    This gives

    f(x)=p(x)(x-\alpha)+r(x)

    but since

    (x - \alpha)

    Is degree 1 r(x) must be degree zero or a constant this gives


    f(x)=p(x)(x-\alpha)+R

    Now just evaluate at alpha

    f(\alpha)=p(x)(\alpha-\alpha)+R=R

    So

    (x - \alpha)|f(x) \iff f(\alpha)=0
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  3. #3
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    for any polynomial we can write:

    f(x) - f(a) = (x - a)q(x) (see the previous post).

    if f(x) = (x - a)k(x), we get f(a) = (x - a)(k(x) - q(x)).

    the LHS is a constant polynomial, and the RHS is a polynomial of degree ≥ 1, which is a contradiction....unless k(x) = q(x),

    in which case we have f(a) = 0.
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