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Polynomial division

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]g(x) = x^3 + 1[/tex] divides [tex]f(x) = x^{9999} +1[/tex].



    2. Relevant equations



    3. The attempt at a solution

    [tex]g(x)[/tex] can obviously be factored into the irreducible polynomials [tex](x+1)(x^2 - x + 1)[/tex] in [tex]Z[x][/tex], and since [tex]f(-1) = (-1)^{9999} + 1 = 0[/tex], the factor theorem gives that [tex](x+1)[/tex] divides [tex]f(x)[/tex].

    Furthermore, we get

    [tex]x^{9999}+1 = (x^2 - x + 1) q(x) + r(x)[/tex]

    where [tex]r(x) = Ax+B[/tex] since [tex]deg(r(x)) < deg(x^2 - x + 1)[/tex] if [tex]r(x) \neq 0[/tex].

    So, showing that [tex]A = B = 0[/tex] would be a good idea, which I have failed to do throughout past trials. I suspect there's an "obvious", clever trick to this, but I'm currently not seeing it.

    Another approach would probably be to use [tex]x^{9999}+1 = (x+1)(x^2 - x + 1) q_{2}(x) + r_{2}(x)[/tex] where [tex]r_{2}(x) = Cx^2 + Dx + E[/tex], and so, [tex]x = -1[/tex] yields [tex]C - D + E = 0[/tex], but that hasn't gotten me anywhere either.


    Note: I'm assuming that I'm not supposed to use complex roots to factor [tex]x^2 - x + 1[/tex], but the problem doesn't specify that such an assumption is necessary.
     
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 29, 2010 #2

    Mark44

    Staff: Mentor

    You've shown that x + 1 is a factor by showing that f(-1) = 0. The other two factors of x^3 + 1 are x = 1/2 +/- i sqrt(3)/2. If you write these in polar form it's pretty easy to raise them to the 9999th power, and thus show that f(1/2 +/- i sqrt(3)/2) = 0.
     
  4. Apr 29, 2010 #3

    Ugh, yay for assumptions! Thank you!
     
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