Factorize

  • Thread starter The legend
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  • #1
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Homework Statement



Factor the equation (without complex numbers)

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

This is a olympiad question

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!
 

Answers and Replies

  • #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
 
  • #3
2,967
5
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.
 
  • #4
424
0
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.
Sorry, but i dont know this part of mathematics. Can you please suggest some place(website etc) where i can learn it by myself?
 
  • #5
2,967
5
How do you expect to go to a Mathematics Olympiad and don't know about complex numbers?
 
  • #6
424
0
complex numbers are not in the syllabus of the olympiad. Thats why i mentioned it in the question.
 
  • #7
424
0
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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