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My question is:

Factorize: $$x^{11} + x^{10} + x^9 + \cdots + x + 1$$

Any help to solve this question would be greatly appreciated.

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    Over which field?2012-07-07
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    So far I'm the only person who's up-voted this question.2012-07-07

2 Answers 2

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$$ \begin{align} & {}\quad (x^{11} + x^{10}) + (x^9 + x^8)+(x^7+x^6)+(x^5+x^4)+(x^3+x^2 )+( x + 1)\\[8pt] & =x^{10}(x+1)+x^8(x+1)+x^6(x+1)+x^4(x+1)+x^2(x+1)+(x+1)\\[8pt] & =(x+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\[8pt] & =(x+1)(x^8(x^2+1)+x^4(x^2+1)+x^2+1)\\[8pt] & =(x+1)((x^2+1)(x^8+x^4+1))\\[8pt] & =(x+1)(x^2+1)(x^4+1-x^2)(x^4+1+x^2)\\[8pt] & =(x+1)(x^2+1)(x^4+1-x^2)(x^2+1-x)(x^2+1+x) \end{align} $$

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    @JyrkiLahtonen:But? Any suggestion for the above solution?2012-07-07
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    @JyrkiLahtonen: Umm, yes. It can go further.2012-07-07
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    Don't worry about it too much. Anyone who has seen cyclotomic polynomials will recognize this, and know that it will factor further. The question "how" is then also answered to an extent.2012-07-07
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    +1 I like your answer, because it shows a Gauss like way of grouping things, to simplify the problem.2012-07-07
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Since $x^{11}+x^{10}+\ldots + x+1 = \frac{x^{12}-1}{x-1}$ we may first factorize $x^{12}-1$ and then divide by the factor $x-1$: \begin{align*} x^{12}-1 &= (x^6-1)(x^6+1)\\ &= (x^3-1)(x^3+1)(x^6+1)\\ &=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1), \end{align*} hence $$x^{11}+x^{10}+\ldots +x+1 = (x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1).$$ It is an easy exercise to show that the factors are irreducible over $\mathbb Q$. In fact, the factors are the cyclotomic polynomials of the divisors of 12 (except 1).

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    What does this mean : the factors are the cyclotomic polynomials of the divisors of 12 (except 1).2012-07-07
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    Like chemistry! You introduced the catalyst $x-1$, it helped our polynomial to break down, then you recovered the catalyst at the end.2012-07-07
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    @meg_1997 See http://en.wikipedia.org/wiki/Cyclotomic_polynomials. In general, $X^n-1$ has the factorization $X^n-1 = \prod_{d|n} \Phi_d$. In this case, $X^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$. Dividing by $\Phi_1 = X-1$ gives the factorization of $1+X+\ldots+x^{11}$.2012-07-07