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I want to factorize this polynomial $x^{7}+1$.

The result that I expect is $(x+1)(x^{3}+x+1)(x^{3}+x^{2}+x+1)$

What is the best way to proceed?

As it seems the factorization is conducted in $\mathbb F_2$.

user
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baskon1
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    Why do you expect that? – Angina Seng Feb 04 '19 at 18:51
  • $(1 + x) (1 - x + x^2 - x^3 + x^4 - x^5 + x^6)$. – David G. Stork Feb 04 '19 at 18:53
  • It is part of a polynomial coding problem. I was given that this polynomial is equal to that factorised polynomial. However I dont get how they got this result, and I dont want to take it for granted... Also we need 3rd degree factors for the polynomial generator.. I think that the 6th degree (Thanks David G. Stork) can be analyzed further.. Will try this.. – baskon1 Feb 04 '19 at 18:55
  • What is the coefficient ring for your polynomials? Mayby your factorization holds over $\Bbb Z_2$? I haven't checked yet . . . – Robert Lewis Feb 04 '19 at 18:56
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    The factorization you have given is incorrect, both over $\Bbb{Q}$ and over any finite field. – Servaes Feb 04 '19 at 18:58
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    Over $\Bbb F_2$, the factorisation is $(x+1)(x^3+x+1)(x^3+x^2+1)$. – Angina Seng Feb 04 '19 at 19:00
  • Lord Shark the Unknown I think you are correct. Since this code is binary, the terms will get eliminated .. It now makes sense! – baskon1 Feb 04 '19 at 19:06

2 Answers2

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$$x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=$$ $$=(x+1)(x^6+x^5+x^4+x^3+x^2+x+1)=$$ $$=(x+1)(x^6+x^4+x^3+x^5+x^3+x^2+x^3+x+1)=$$ $$=(x+1)(x^3(x^3+x+1)+x^2(x^3+x+1)+x^3+x+1)=$$ $$=(x+1)(x^3+x+1)(x^3+x^2+1).$$ Can you end it now?

Michael Rozenberg
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The polynomial $x^7+1$ has a factor $x+1$, which can easily be seen from the fact that $(-1)^7+1=0$. The remainder is the $14$th cyclotomic polynomial, which is irreducible as all cyclotomic polynomials are.

Servaes
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  • But wasn’t OP dealing with $\Bbb F_2$-polynomials only? – Lubin Feb 04 '19 at 23:05
  • Unless I'm missing something, this isn't right. $\Phi_n$ is irreducible over $\mathbb Q$, but as Michael Rozenberg's answer shows, $\Phi_{14}$ is not irreducible over $\mathbb F_2$. – YiFan Tey Feb 05 '19 at 03:40
  • @YiFan Indeed it isn't irreducible over $\Bbb{F}_2$, but there was no mention of any field in the original question. – Servaes Feb 05 '19 at 07:14
  • @Servaes I see, I didn't check the edit history. My bad. – YiFan Tey Feb 05 '19 at 09:36