Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$ When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say 
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{cx+d}$$
The correct way is 
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$
 A: Because 
$$\frac{B}{cx+d}+ \frac{C}{cx+d} = \frac{B+C}{cx+d}\ne\frac{1}{(cx+d)^2}$$ for any $B$ and $C$.
A: HINT $\rm\displaystyle\quad \frac{f(x)}{g(x)\ (c\:x+d)^2}\: =\: \frac{h(x)}{(a\:x+b)\:(c\:x+d)}\ \ \Rightarrow\ \ f(x)\:(a\:x+b)\: =\: g(x)\:h(x)\:(c\:x+d)\:$  
hence, evaluating at $\rm\:x = -d/c\:$ yields that either $\rm\:f(-d/c) = 0\:$ (so the LHS isn't in lowest terms) or $\rm\ a\:x+b\ $ has root $\rm\:x = -d/c\:$ (so the RHS has denominator $\rm\:(c\:x+d)^2\:$ times constant).
NOTE $\ $ This is simply a rational function analog of the fact that rational numbers have unique prime factorization. Hence above, if the LHS is in lowest terms and the  prime $\rm\:c\:x+d\:$ occurs to power $\:2\:$ in its denominator, then the same must be true of the RHS when in lowest terms. Therefore $\rm\ a\:x+b\: =\: e\:(c\:x+d)\:$ for some constant $\rm\:e\:.$
A: The degree of the denominator is 2, so the numerator has to be of degree 1. So you can either assume it to be $Bx+C$ or (better still), $B(bx+c)+C$, so that 
$\frac{B(bx+c)+C}{(bx+c)^2} = \frac{B}{bx+c} + \frac{c}{(bx+c)^2}.$
This generalizes to the case when the denominator has degree $n$. 
