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Proof that a polynomial is a factor

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that [itex]x+a[/itex] is a factor of [itex]x^{n}+a^{n}[/itex]for all odd n.

    3. The attempt at a solution

    (1) Assume that [itex]x+a[/itex] is a factor of [itex]x^{n}+a^{n}[/itex]for all odd n. This implies that when [itex]x^{n}+a^{n}[/itex] is divided by [itex]x+a[/itex] the remainder is zero.

    I don't know - is this a sensible 1st step?

    What to do next?
     
  2. jcsd
  3. Mar 22, 2012 #2
    I'd go for contradiction, suppose it isn't and see what happens
     
  4. Mar 22, 2012 #3
    Could you give me some hint what to do next? I really don't understand how to incorporate the "for all odd n" part into the reasoning process.
     
  5. Mar 22, 2012 #4
    What property do things with odd powers have that things with even powers don't?
     
  6. Mar 22, 2012 #5

    tiny-tim

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    hi mindauggas! :smile:

    hint: roots? :wink:
     
  7. Mar 22, 2012 #6

    HallsofIvy

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    As tiny tim suggests, x- a is a factor of polynomial p(x) if and only if p(a)= 0. Of course, x+ a= x-(-a).
     
  8. Mar 22, 2012 #7
    Probably a lot of properties (e.g. the "thing" with odd power is not always positive after raising it to the power and the "thing" with even powers is always positive ("thing" in ℝ)).

    Presumably you are asking the most general property that distinguishes them ... this I beg you to tell me, please (this would help me not only in this problem but in general).
     
  9. Mar 22, 2012 #8
    Yes, you've got it!
    How can you use that to arrive at a contradiction now?
     
  10. Mar 22, 2012 #9
    So I just put - [itex]f(-a)=(-a)^{n}+a^{n}=0[/itex], under the condition that n is odd (now I see how that came into play) and that's it?

    Thanks.
     
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