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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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2  
Over which field? –  Did Jul 7 '12 at 20:35
    
So far I'm the only person who's up-voted this question. –  Michael Hardy Jul 7 '12 at 22:20

2 Answers 2

up vote 6 down vote accepted

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

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@JyrkiLahtonen:But? Any suggestion for the above solution? –  meg_1997 Jul 7 '12 at 20:37
    
@JyrkiLahtonen: Umm, yes. It can go further. –  Gigili Jul 7 '12 at 20:38
    
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. –  Jyrki Lahtonen Jul 7 '12 at 20:40
    
+1 I like your answer, because it shows a Gauss like way of grouping things, to simplify the problem. –  draks ... Jul 7 '12 at 20:57
    
+1 I like your answer too. :) –  Babak S. Jul 9 '12 at 13:11

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). –  meg_1997 Jul 7 '12 at 20:46
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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. –  André Nicolas Jul 7 '12 at 20:47
    
+1 for the answer and the chemistry comment! –  draks ... Jul 7 '12 at 20:50
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@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 $1+X+\ldots+x^{11}$. –  marlu Jul 7 '12 at 20:53

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