The method of grouping terms can be used to factor all-one polynomials, i.e. polynomials of the form $$ \sum_{m=0}^{n-1} x^m $$ when $n$ is composite. (When $n$ is prime, these polynomials are irreducible, so there is nothing to do in that case.)

Let us consider a few examples:

$n = 4$ \begin{eqnarray*} 1 + x + x^2 + x^3 = \\ (1 + x) + (x^2 + x^3) = \\ (1 + x) + x^2 (1 + x) = \\ (1 + x) (1 + x^2) \end{eqnarray*} $n = 6$ \begin{eqnarray*} 1 + x + x^2 + x^3 + x^4 + x^5 = \\ (1 + x + x^2) + (x^3 + x^4 + x^5) = \\ (1 + x + x^2) + x^3 (1 + x + x^2) = \\ (1 + x^3) (1 + x + x^2) \end{eqnarray*} $n = 8$ \begin{eqnarray*} 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = \\ (1 + x + x^2 + x^3) + (x^4 + (x^5 + x^6 + x^7) = \\ (1 + x + x^2 + x^3) + x^4 (1 + x + x^2 + x^3) =\\ (1 + x^4) (1 + x + x^2 + x^3) \end{eqnarray*}Combining this result with the factorization we have for the case $n=4$ we obtain the following: \begin{eqnarray*} 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = \\ (1 + x) (1 + x^2) (1 + x^4) \end{eqnarray*} $n = 9$ \begin{eqnarray*} 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 = \\ (1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7 + x^8) = \\ (1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) = \\ (1 + x + x^2) (1 + x^3 + x^6) \end{eqnarray*} $n = 12$ \begin{eqnarray*} 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} + x^{11} = \\ (1 + x + x^2) + (x^3 + x^4 + x^5) + (x^6 + x^7 + x^8) + (x^9 + x^{10} + x^{11}) = \\ (1 + x + x^2) + x^3 (1 + x + x^2) + x^6 (1 + x + x^2) + x^9 (1 + x + x^2) = \\ (1 + x + x^2) (1 + x^3 + x^6 + x^9) = \\ (1 + x + x^2) ((1 + x^3) + (x^6 + x^9)) = \\ (1 + x + x^2) ((1 + x^3) + x^6 (1 + x^3)) = \\ (1 + x + x^2) (1 + x^3) (1 + x^6) \end{eqnarray*} It might be worth pointing out that the polynomials produced by this factorization are not all irreducible. For instance, $$ 1 + x^3 = (1 + x) (1 - x + x^2). $$ However, to obtain this factorization, one needs to use some techique other than the grouping method. Likewise. the polynomial $1 + x^6$ is also reducible.