factoring
posted by alisia* .
could anybody help me figure out step by step how to factor this expression:
x^10+x^5+1(please)
p.s. i need to explain this on the hw

factoring 
mktintin
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 
mktintin
so do u succeed??

factoring 
Count Iblis
As mktintin 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 = (xw1)(xw2)...(xw10)
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 [1sqrt(5)] + 1/4 sqrt{2/3 [5 + sqrt(5)]}
cos(2/3 pi) =  1/2
cos(16/15 pi) = 1/8 [1sqrt(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
ok, i think i get it now. thank you!

factoring 
alicia*
i got it. thanks a lot*
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