Polynomial $x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ is reducible over $\mathbb{Q}$?

Question

Clearly, there are no roots, but how can I find factors of higher degree?

Answer 1

To see how to get the answer, know the factorization of a finite geometric series:

$$ \begin{array}{ll} 1+x+\cdots+x^8 & \displaystyle=\frac{x^9-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3)^3-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3-1)\big((x^3)^2+x^3+1\big)}{x-1} \\[5pt] & = (x^2+x+1)(x^6+x^3+1) \end{array} $$

As Cauchy mentions in the comments, this generalizes with cyclotomic polynommials.

Answer 2

This is $$(x^6+x^3+1)(x^2+x+1).$$

In fact a polynomial $$\sum_{j=0}^m x^j=x^m+x^{m-1}+\cdots+x+1$$ for $m\ge1$ is irreducible over $\Bbb Q$ iff $m+1$ is prime.

Answer 3

A good way to see this, if you don't know about geometric series or about cyclotomic polynomials, is to factor by grouping. Start by splitting the nine terms into three groups of three:

$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1\\ = (x^8+x^7+x^6)+(x^5+x^4+x^3)+(x^2+x+1)$$

Now, factor out a GCF from each group:

$$=x^6(x^2+x+1)+x^3(x^2+x+1)+1(x^2+x+1)$$

Finally, pull out the common factor of $(x^2+x+1)$:

$$=(x^6+x^3+1)(x^2+x+1)$$

The trick that made this work was that the number of terms, $9$, was composite, allowing us to make regular groups.

Answer 4

Hint $\ $ The sequence $\rm\:F_{\large n} = (x^n\!-\!1)/(x\!-\!1)\:$ is a divisibility sequence, $ $ i.e. $\,\rm m\mid n\,\Rightarrow\, F_{\large m}\mid F_{\large n}$

Hence, $ $ in particular $\rm \, F_{\large 3}\mid F_{\large 9}\,$ so it has a factor $\,\rm F_{\large 3} = (x^3\!-\!1)/(x\!-\!1) = x^2+x+1$

Remark $ $ Furthermore, $\rm\,F_{\large n}\,$ is a strong divisibility sequence, i.e. $\rm\:(F_{\large m},F_{\large n}) = F_{\large (m,n)},\:$ where $\,\rm (m,n):= \gcd(m,n).\,$ This implies an intimate relationship between divisibility properties of the $\rm\:F_{\large n}\:$ and integers $\rm\:n\:$ (so, in particular, relation between notions associated with divisibility, such as irreducible = prime).

You may recognize these properties from another common strong divisibility sequence - the Fibonacci numbers. They share many properties, e.g. an addition law below (from this answer)

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Regarding $\,f_n = \dfrac{x^n-y^n}{x-y},\,$ which satisfies $\,f_{n+2} = (x\!+\!y) f_{n+1}-xy\, f_n,\,$ a similar proof shows that it satisfies the addition law $\, f_{m+n} = f_{m+1} f_n - xy\, f_m f_{n-1},\ $ i.e.

$$ \dfrac{x^{m+n}\!-y^{m+n}}{x-y}\,=\, \dfrac{x^{m+1}\!-y^{m+1}}{x-y}\,\dfrac{x^{n}\!-y^{n}}{x-y} - xy\, \dfrac{x^{m}-y^{m}}{x-y}\,\dfrac{x^{n-1}-y^{n-1}}{x-y}$$