The length of the segment between the points $(2a, a-4)$ and $(4, -1)$ is $2\sqrt{10}$ units. What is the product of all possible values for $a$?
$(2a-4)^2+(a-3)^2=40$
$4a^2-16a+16+a^2-6a+9=40$
$5a^2-22a+25=40$
$5a^2-22a-15=0$.
The product is $-\frac{15}{5}=\boxed{-3}$ by Vieta.