Question 854555
<pre>
{{{(3x)/(x-3)^2}}}

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.

{{{3x/(x-3)^2}}}{{{""=""}}}{{{A/(x-3)^2+B/(x-3)}}}

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:

{{{3x}}}{{{""=""}}}{{{A+B(x-3)}}}

{{{3x}}}{{{""=""}}}{{{A+Bx-3B}}}

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

{{{3x}}}{{{""=""}}}{{{Bx}}}

So B = 3

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

{{{"0"}}}{{{""=""}}}{{{A-3B}}}

Substitute 3 for B

{{{"0"}}}{{{""=""}}}{{{A-3(3)}}}

{{{"0"}}}{{{""=""}}}{{{A-9}}}

{{{9}}}{{{""=""}}}{{{A}}}

So

{{{3x/(x-3)^2}}}{{{""=""}}}{{{A/(x-3)^2+B/(x-3)}}}

becomes

{{{3x/(x-3)^2}}}{{{""=""}}}{{{9/(x-3)^2+3/(x-3)}}}

Edwin</pre>