We find \(B\) using Heaviside's method:
\[ \dfrac{x-4}{(x-2)^2}\cdot (x-2)^2=x-4 \] \[ B=x-4\Huge|_{\normalsize x=2}\normalsize=-2 \] \[ \dfrac{x-4}{(x-2)^2}=\dfrac{A}{x-2}-\dfrac{2}{(x-2)^2} \]Clearing the fractions gives \(x-4=A(x-2)-2=Ax-2(A+1)\). So \(A=1\).
Thus the decomposition is
\[ \dfrac{x-4}{(x-2)^2}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2} \]