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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2$\begingroup$ Over which field? $\endgroup$– DidCommented Jul 7, 2012 at 20:35
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$\begingroup$ So far I'm the only person who's up-voted this question. $\endgroup$– Michael HardyCommented Jul 7, 2012 at 22:20
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2 Answers
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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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$\begingroup$ @JyrkiLahtonen:But? Any suggestion for the above solution? $\endgroup$– mghCommented Jul 7, 2012 at 20:37
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$\begingroup$ @JyrkiLahtonen: Umm, yes. It can go further. $\endgroup$– GigiliCommented Jul 7, 2012 at 20:38
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$\begingroup$ 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. $\endgroup$ Commented Jul 7, 2012 at 20:40
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$\begingroup$ +1 I like your answer, because it shows a Gauss like way of grouping things, to simplify the problem. $\endgroup$ Commented Jul 7, 2012 at 20:57
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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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$\begingroup$ What does this mean : the factors are the cyclotomic polynomials of the divisors of 12 (except 1). $\endgroup$– mghCommented Jul 7, 2012 at 20:46
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6$\begingroup$ Like chemistry! You introduced the catalyst $x-1$, it helped our polynomial to break down, then you recovered the catalyst at the end. $\endgroup$ Commented Jul 7, 2012 at 20:47
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1$\begingroup$ @meg_1997 See 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 $1+X+\ldots+x^{11}$. $\endgroup$– marluCommented Jul 7, 2012 at 20:53