$\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ when resolved into partial fractions is equal to?
I solved it using $\frac{23x - 11x^2}{(2x-1)(9-x^2)} = \frac{A}{2x-1} + \frac{B}{3-x} + \frac{C}{3+x}$, but this is very long method to solve, as after comparing I will get three variables equations. Please tell me any shorter method to do it.
$$\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{23x-11x^2}{(2x-1)(3-x)(3+x)}$$
The denominator of $A$ vanishes when $x=\frac12$, omit the vanishing term, we have
$$A=\frac{23\left( \frac12\right)-11\left( \frac12\right)^2}{(3-\frac12)(3+\frac12)}$$
The denominator of $B$ vanishes when $x=3$, again, we have
$$B = \frac{23(3)-11(3)^2}{(2(3)-1)(3+3)}$$
Similarly for $C$.
The main idea is
$$23x-11x^2 = A(3-x)(3+x)+B(2x-1)(3+x)+C(2x-1)(3-x)$$
By using the terms that the other two terms vanishes, we can solve a variable easily.
For example, by letting $x=\frac12$, the last two terms on the right vanishes and we are left with
$$23\left( \frac12\right)-11\left( \frac12\right)^2=A\left( 3-\frac12\right) \left( 3+\frac12\right)$$
This is known as Heaviside cover-up method. Refer to case $2$ of the wikipedia if the factors of the denominator include powers of one expression