jmc

First, we factor the denominator, to get $$\frac{x^2+5x+\alpha }{x^2+7x-44}=\frac{x^2+5x+\alpha }{(x-4)(x+11)}.$$If this fraction can be expressed as a quotient of two linear functions, then the numerator must have either a factor of $x-4$ or $x+11$.

If the numerator has a factor of $x-4$, then by the factor theorem, it must be 0 when $x=4$. Hence, $4^2+5\cdot 4+\alpha =0$, which means $\alpha =-36$.

If the numerator has a factor of $x+11$, then it must be 0 when $x=-11$. Hence, $(-11{)}^2+5\cdot (-11)+\alpha =0$, which means $\alpha =-66$.

Therefore, the sum of all possible values of $\alpha$ is $-36+(-66)=\boxed{-102}$.