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I'm getting ready for the exam in algebra. And I have problem that I can't solve again.

I'm understand what I need to do if I have a polynomial with a given degree. But I very confused by arbitrary degree in this problem.

Given polynomial $$x^{2n}-2x^n+2$$ The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$.

Thanks in advance.

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up vote 3 down vote accepted

$$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$

Now note that: $$x^n-2^{0.5}e^{\frac{-i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$ $$x^n-2^{0.5}e^{\frac{i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})$$

Hence: $$x^{2n}-2x^n+2=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-(2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}}+2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})x+2^{\frac{1}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-2(2^{\frac{1}{2n}}\cos({\frac{\pi}{4}+\frac{2\pi k}{n}}))x+2^{\frac{1}{n}})$$

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It should be $\displaystyle{\prod_{k=1}^n\left(x^2-2\left( 2^{ \frac{1}{2n} } \cos \left( \dfrac{\pi}{4}+\dfrac{2\pi k}{n} \right)\right)x+2^{\frac{1}{n}}\right)}$, right? –  P.. Jan 13 '13 at 11:25
    
@Pambos Yes thank you –  Amr Jan 13 '13 at 11:33
    
@Amr Thank you! –  Oiale Jan 13 '13 at 12:37
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