Question 580852
The only way I can think of factoring it is:
{{{1+x+x^2+x^3+x^4+x^5=(x^6-1)/(x-1)=(x^3-1)(x^3+1)/(x-1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)/(x-1)=(x^2+x+1)(x+1)(x^2-x+1)}}}
and there is no further factoring because the quadratic polynomials have no zeros:
{{{x^2+x+1>x^2+x+1/4=(x+1/2)^2>=0}}} and
{{{x^2-x+1>x^2-x+1/4=(x-1/2)^2>=0}}}