Polynomial $x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ is reducible over $\mathbb{Q}$? Clearly, there are no roots, but how can I find factors of higher degree?

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.
