jmc

We can factor $x$ out of the numerator so that we're left with $$\frac{x(x^2-2x-8)}{x+2}=\frac{x(x-4)(x+2)}{x+2}$$After canceling out the $x+2$ from the numerator and denominator, we have $x(x-4)=5$. Solving for the roots of a quadratic equation, we have $x^2-4x-5=0$, which gives us $(x-5)(x+1)=0$ and $x=5$ or $x=-1$. The sum of these values is $\boxed{4}$, which is our answer.

Alternatively, since the sum of the solutions for a quadratic with the equation $a{x}^2+bx+c=0$ is $-b/a$, the sum of the zeros of the quadratic $x^2-4x-5$ is $4/1=\boxed{4}$.