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?

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.
