To factorise x^5 + x + 1?

You want x^5 + x + 1 = 0, or x^5 + x = -1. Maybe we can make x^5 and x complex conjugates with real parts -0.5--that would give us two roots. Taking x = re(it), x^5 = re^(5it) = re(-it), so 5it == -it (mod 2pi), so 6t = 2pi (say), so t = pi/3. Requiring x to have real part -0.5 then forces rcos(t) = -0.5 = r*0.5, so r=-1.

So x=-e^(i pi/3) and -e^(-i pi/3) are roots. These are themselves roots of the quadratic (x + e^(i pi/3))(x + e^(-i pi/3)) = x^2 + x + 1. You can perform polynomial division to get (x^5 + x + 1)/(x^2 + x + 1) = x^3 - x^2 + 1. This polynomial is irreducible: consider it as an element of Z/2Z = GF(2), the finite field on 2 elements. Then it is x^3 + x^2 + 1, and if it is reducible, it must have a linear factor. But plugging in 0 and 1 directly, we get 0+0+1 = 1 and 1+1+1 = 1, so these can't be roots, and it can't have a linear factor. Since this polynomial is irreducible over a quotient field, it is irreducible over the rationals. In all, the full factorization over the rationals is

x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1).

(Note: I had Wolfram Alpha solve the roots and reverse engineered a possible solution, so this method was not obvious.)

There are the two ways I came up with for this problem:

x(x^4+1)+1

(x^2+x+1)(x^3-x^2+1)

Hope I helped!