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I need to find a polynomials product that give me $x^6+1$ at $\mathbb{R}[X]$ and at $\mathbb{C}[X]$.

I need that the product will be of irreducible polynumials...

Thank you!

CS1
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2 Answers2

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You may like $$\begin{align}x^6+1&=(x^2+1)(x^4-x^2+1)\\&=(x^2+1)(x^2+\sqrt 3x+1)(x^2-\sqrt 3x+1)\\&=(x+i)(x-i)(x^2+\sqrt 3x+1)(x^2-\sqrt 3x+1).\end{align}$$

mathlove
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Use the fact every complex root of Polynomials with real coefficients comes with its conjugate.

Then you can find the roots of 1 by using:

$e^{\pi+2ki}=-1$ for all integers k and ${e^{\pi}} ^{1/6}=e^{\pi/6+1/3 k\pi i}$

b00n heT
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  • You can help me with this? How do I get from this a irreducible polynumial? – CS1 Jan 09 '14 at 18:18
  • Using Fundamental theorem of Algebra you know that every polynomial of order n has n roots (with multiplicity) in $\mathbb{C}[X]$ so you know you can divide that polynomial in a multiplication of 6 monoms, each of the form $(x-x_0)$ as shown in the comments above after having found $(x^2+1)$ you do common polynomial division and the use formula for solving polynomials of the form $a{x^d}^2+bx^d+c=0$ At last you go back to $\mathbb{R}[X]$ by combining the conjugate complex roots – b00n heT Jan 09 '14 at 18:28
  • Yes, but it different section, I need to find all of them at $\mathbb{C}$... At the comment above is good for $\mathbb{R}$. Thank you!!! – CS1 Jan 09 '14 at 18:30
  • How do I use the $e^{...}$ method to find the polynomials, this what I meant :-). – CS1 Jan 09 '14 at 18:32