Factoring $x^5 + x + 1$

As Gerry suggests, we can add and subtract $\;+ x^2 - x^2 = 0 $ without changing the expression:

$$\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}$$

Now factor out the common factor...which gives us $$\Big(x^2(x-1) + 1\Big)(x^2 + x + 1) = (x^3 - x^2 + 1)(x^2 + x + 1)$$

You can verify that both primitive $3$rd roots of unity are roots of this equation, since $5$ is $2$ mod $3$. This means the corresponding cyclotomic polynomial, $x^2+x+1$, divides $x^5+x+1$, and you can do polynomial division to find the other factor.

The answer is $(x^2+x+1)(x^3-x^2+1)$.