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August 23, 2014

August 23, 2014

Posted by **alisia*** on Monday, February 11, 2008 at 6:04pm.

x^10+x^5+1(please)

p.s. i need to explain this on the hw

- factoring -
**mk-tintin**, Monday, February 11, 2008 at 6:10pmtry 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 -
**mk-tintin**, Monday, February 11, 2008 at 6:19pmso do u succeed??

- factoring -
**Count Iblis**, Monday, February 11, 2008 at 8:35pmAs 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 -
**JSS**, Monday, February 11, 2008 at 9:30pmok, i think i get it now. thank you!

- factoring -
**alicia***, Monday, February 11, 2008 at 9:32pmi got it. thanks a lot*

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