How to divide it: $\frac{cx+d}{ax+b}$?

Question

Somebody divided $\frac{cx+d}{ax+b}$ into $$ \frac{c}{a} + \frac{d- \frac{bc}{a}}{ax+b} .$$ For use for integrals.

Does anybody knows how was it done? Could you show me how to do something like that, please?

Thanks in advance.

Answer 1

It's polynomial long division. In this relatively simple case (first degree polynomials in both numerator and denominator) you can also do it like this, using an "add-and-subtract" trick: $$ \frac{cx+d}{ax+b} = \frac{c(x + d/c)}{a(x+b/a)} = \frac{c}{a} \frac{(x+b/a)+(d/c-b/a)}{x+b/a} = \frac{c}{a} \left( 1 + \frac{d/c-b/a}{x+b/a} \right).$$

Answer 2

It is the linear case of the polynomial remainder theorem: $\rm\ f(x)\ \ mod\ \ a\:x + b\ =\ f(-b/a)\:.\ $ This has a very simple proof. Namely, by the division algorithm for polynomials over a field, $\rm\ f(x)\ =\ (a\ x + b)\ q(x) + r\ $ so $\rm\ r = f\:(-b/a)\ $ follows by evaluation at $\rm\ x = -b/a\:.\ $ In your case $\rm\ f(x)\ =\ c\ x + d\ $ so $\rm\ r = f\:(-b/a) = d - bc/a\:.\ $ Also, comparing degrees shows that the quotient has degree zero $\rm\:q(x) = q\:.\ $ Comparing leading coefficients yields $\rm\ c = a\ q\ $ so $\rm\ q(x) = q = c/a\:.\ $

Hence $\rm\displaystyle\quad c\ x + d\ =\ (a\ x + b)\ q + r\ \ \Rightarrow\ \ \frac{c\ x + d}{a\ x + b}\ =\ q + \frac{r}{a\ x + b},\quad q = \frac{c}a,\ \ r = d-\frac{bc}a$

More generally, over any Euclidean domain, i.e. any integral domain with a division algorithm, one may use the division algorithm to decompose a fraction into an integral and fractional part. This is frequently the inlined $\:$ descent $\: $ step in proofs that implicitly invoke unique factorization.$\ $ For example, see my analysis of a proof of square-root irrationality presented by John Conway.