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Factorize

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Factor the equation (without complex numbers)

    [tex]
    a^{10} + a^{5} + 1
    [/tex]

    This is a olympiad question

    3. The attempt at a solution

    I substituted [tex]a^{5} = x[/tex] getting a quadratic eqation. But when I factored the quadratic equation I get complex roots and this is against the question.

    Please help!
     
  2. jcsd
  3. Sep 2, 2010 #2
    well, using your substitution:
    x^2 + x + 1

    I don't know if this is the final answer they are looking for but you could add a 'clever' version of 0 to simplify it a bit
     
  4. Sep 2, 2010 #3
    The solutions of [itex]x^{2} + x + 1 = 0[/itex] are complex numbers:

    [tex]
    x_{1/2} = \frac{-1 \pm i \sqrt{3}}{2}
    [/tex]

    The modulus of these numbers is:

    [tex]
    |x_{1/2}| = 1
    [/tex]

    and their arguments (limited to the interval [itex][0, 2\pi)[/itex]) are:

    [tex]
    \arg{(x_{1})} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
    [/tex]

    [tex]
    \arg{(x_{2})} = \pi + \frac{\pi}{3} = \frac{4\pi}{3}
    [/tex]

    Take the fifth root of each of them and combine the complex conjugate roots in five pairs.
     
  5. Sep 3, 2010 #4
    Sorry, but i dont know this part of mathematics. Can you please suggest some place(website etc) where i can learn it by myself?
     
  6. Sep 3, 2010 #5
    How do you expect to go to a Mathematics Olympiad and don't know about complex numbers?
     
  7. Sep 4, 2010 #6
    complex numbers are not in the syllabus of the olympiad. Thats why i mentioned it in the question.
     
  8. Sep 4, 2010 #7
    I tried out a different approach(without complex nos) and got it!

    Take [tex]a^{10} + a^{5} + 1 = y[/tex] and [tex]a^5 = x[/tex]

    then

    [tex]
    y = x^{2} + x + 1
    [/tex]
    [tex]
    (x - 1) y = (x - 1)(x^{2} + x + 1)
    [/tex]
    [tex]
    (x - 1) y = x^{3} - 1
    [/tex]
    [tex]
    y = \frac{a^(5)3 - 1^{5}}{a^{5} - 1}
    [/tex]

    Now simplifying using (x^5 - y^5) and dividing i get the answer to be

    [tex]
    y = (a^{2} + a + 1)(a^{8} - a^{7} + a^{5} -a^{4} + a^{3} - a + 1)
    [/tex]

    Right , isnt it?


    anyway..Dickfore, can you please suggest a good book or website to self learn complex numbers?
     
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