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could anybody help me figure out step by step how to factor this expression:
p.s. i need to explain this on the hw

  • factoring -

    try to solve this equation with X=x^5
    so u have x^10+x^5+1--->> X²+X+1
    and it's much easier to solve this equation

  • factoring -

    so do u succeed??

  • factoring -

    As mk-tintin explained, you first solve the equation:

    X²+X+1 = 0 --->

    X = -1/2 +- 1/2 sqrt(3) i. We can write this as:

    X = cos(4/3 pi) +- i sin(4/3 pi) (+- sign reversed relative to previous expression)

    This can be written in exponential form as:

    X = exp(+- 4/3 pi i)

    The next step is to compute the five solutions of x^5 = X:

    x = exp(+- 4/15 pi i + 2/5 n pi i)

    for n = 0, 1,...,4 or some other range of five.

    So we have ten solutions in total, let's denote them by w1, w2, ..., w10. Then:

    x^10+x^5+1 = (x-w1)(x-w2)...(x-w10)

    Finally, you want to rewrite the right hand side in terms of real factors only. To do this,
    you exploit the fact that the roots of a polynomial with real coefficients come in pairs
    that are related under complex conjugation. The two factors corresponding to two
    such roots can be multiplied to yield a real quadratic factor. In this case we can use:

    (x - exp(4/15 pi i + 2/5 n pi i))(x - exp(-4/15 pi i - 2/5 n pi i)) =

    x^2 - 2 cos[(4/15 + 2/5 n) pi] x + 1

    So, the factorization can be written as:

    x^10+x^5+1 = (x^2 - 2 cos(4/15 pi) x + 1)(x^2 -2 cos(2/3 pi) x + 1)
    (x^2 - 2 cos(16/15 pi) x + 1)(x^2 - 2 cos(22/15 pi) x + 1)(x^2 - 2 cos(28/15 pi) x + 1)

    Now, we aren't finished yet, because each of the cosine terms can be simplified:

    cos(4/15 pi) = 1/8 [1-sqrt(5)] + 1/4 sqrt{2/3 [5 + sqrt(5)]}

    cos(2/3 pi) = - 1/2

    cos(16/15 pi) = 1/8 [1-sqrt(5)] - 1/4 sqrt{2/3 [5 + sqrt(5)]}

    cos(22/15 pi) = 1/8 [1+sqrt(5)] - 1/4 sqrt{2/3 [5 - sqrt(5)]}

    cos(4/15 pi) = 1/8 [1+sqrt(5)] + 1/4 sqrt{2/3 [5 - sqrt(5)]}

  • factoring -

    ok, i think i get it now. thank you!

  • factoring -

    i got it. thanks a lot*

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