Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
2  
$\frac{B}{cx+d} + \frac{C}{cx+d}=\frac{B+C}{cx+d}$ 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 $cx+d$... –  J. M. Oct 13 '11 at 12:44
    
@J.M. you're right ... didn't think about that ... –  Jiew Meng Oct 13 '11 at 12:53
    
See this answer for a closed formula for partial fraction decomposition. –  Pierre-Yves Gaillard Oct 13 '11 at 13:08

3 Answers 3

up vote 1 down vote accepted

Because $$\frac{B}{cx+d}+ \frac{C}{cx+d} = \frac{B+C}{cx+d}\ne\frac{1}{(cx+d)^2}$$ for any $B$ and $C$.

share|improve this answer
    
Also think about the case $c=1$ and $d=0$.. –  AD. Oct 13 '11 at 13:04

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 $\rm\:e\:.$

share|improve this answer

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 $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.