# Polynomial-related manipulation

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? – 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

\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}
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).
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
@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