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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

$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$

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    $\begingroup$ $\frac{B}{cx+d} + \frac{C}{cx+d}=\frac{B+C}{cx+d}$ is essentially indistinguishable from $\frac{B}{cx+d}$. If you bring together all the fractions in your second "equation", the denominator would take the form $(ax+b)(cx+d)$, which is short by a factor of $cx+d$... $\endgroup$ – J. M. is a poor mathematician Oct 13 '11 at 12:44
  • $\begingroup$ @J.M. you're right ... didn't think about that ... $\endgroup$ – Jiew Meng Oct 13 '11 at 12:53
  • $\begingroup$ See this answer for a closed formula for partial fraction decomposition. $\endgroup$ – Pierre-Yves Gaillard Oct 13 '11 at 13:08
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Because $$\frac{B}{cx+d}+ \frac{C}{cx+d} = \frac{B+C}{cx+d}\ne\frac{1}{(cx+d)^2}$$ for any $B$ and $C$.

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  • $\begingroup$ Also think about the case $c=1$ and $d=0$.. $\endgroup$ – AD. Oct 13 '11 at 13:04
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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 $\rm\:e\:.$

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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 $n$.

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