You must include a term with a denominator
which is a lower power of a denominator.
So since (x-3)² is to the 2nd power, you must
include a term with the lower power (x-3) in
the denominator.



Notice also that even though (x-3)² is degree 2 if
multiplied out, we do not need "Ax+B" for its 
numerator because it is factored as the power of  
only a first degree binomial.

Clear of fractions:





Since the sides must be identical, we can
equate the like terms on both sides:



So B = 3

There are no constant terms on the left, so put 0



Substitute 3 for B







So



becomes



Edwin