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)} $.