jmc

We can factor by pairing $x^5$ and $-x,$ and $-x^2$ and $-1$: $$\begin{array}{rl}{x}^5-x^2-x-1& =(x^5-x)-(x^2+1)\\ & =x(x^4-1)-(x^2+1)\\ & =x(x^2+1)(x^2-1)-(x^2+1)\\ & =(x^2+1)(x^3-x-1).\end{array}$$If $x^3-x-1$ factors further, then it must have a linear factor, which means it has an integer root. By the Integer Root Theorem, the only possible integer roots are $\pm 1,$ and neither of these work, so $x^3-x-1$ is irreducible.

Thus, $(x^2+1)(x^3-x-1)$ is the complete factorization. Evaluating each factor at 2, we get $(2^2+1)+(2^3-2-1)=\boxed{10}.$