jmc

Factoring the denominator on the left side gives $$\frac{4x}{(x-5)(x-3)}=\frac{A}{x-3}+\frac{B}{x-5}.$$Then, we multiply both sides of the equation by $(x-3)(x-5)$ to get $$4x=A(x-5)+B(x-3).$$If the linear expression $4x$ agrees with the linear expression $A(x-5)+B(x-3)$ at all values of $x$ besides 3 and 5, then the two expressions must agree for $x=3$ and $x=5$ as well. Substituting $x=3$, we get $12=-2A$, so $A=-6$. Likewise, we plug in $x=5$ to solve for $B$. Substituting $x=5$, we get $20=2B$, so $B=10$. Therefore, $(A,B)=\boxed{(-6,10)}.$