text: # Decomposition of polynomial of n-degree into irreducible polynomials.
  
  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 $\mathbb{R}$.
  
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  $$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$
  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: $$x^{2n}-2x^n+2=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-(2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}}+2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})x+2^{\frac{1}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-2(2^{\frac{1}{2n}}\cos({\frac{\pi}{4}+\frac{2\pi k}{n}}))x+2^{\frac{1}{n}})$$
  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
explanation: The text extract is a detailed solution to a math problem. It starts with a problem statement asking for the decomposition of a given polynomial into irreducible polynomials over the real numbers. The solution is provided step-by-step, showing the factorization process. The extract also includes a comment verifying the final form of the solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Decomposition of polynomial of n-degree into irreducible polynomials.
  
  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 $\mathbb{R}$.
  
  -
  
  $$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$
  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: $$x^{2n}-2x^n+2=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-(2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}}+2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})x+2^{\frac{1}{n}})$$ $$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-2(2^{\frac{1}{2n}}\cos({\frac{\pi}{4}+\frac{2\pi k}{n}}))x+2^{\frac{1}{n}})$$
  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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a detailed solution to a math problem. It starts with a problem statement asking for the decomposition of a given polynomial into irreducible polynomials over the real numbers. The solution is provided step-by-step, showing the factorization process. The extract also includes a comment verifying the final form of the solution.
  
  Label: question-and-answer