Thread: solving partial fractions manually

is this correct?

$\displaystyle \frac{8z^2}{(2z-1)(4z-1)} = \frac{A}{2z-1} + \frac{B}{4z-1}$

answer from wolfram is
$\displaystyle \frac{-1}{4z-1}+\frac{2}{2z-1}+1$
how did they get the +1?

You can only do partial fractions when the degree of the numerator is less than the degree of the denominator.

But here, the numerator and denominator both have degree 2.

So long divide first, then use partial fractions on the remainder.

The correct set-up is:

$\displaystyle \displaystyle \frac{8z^2}{(4z-1)(2z-1)} = \frac{A}{4z-1}+\frac{B}{2z-1}+C.$

The denominator is $\displaystyle 8z^2- 6z+ 1$. Clearly the leading "$\displaystyle 8z^2$" will divide into the numerator exactly once.