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$\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ 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.

sawan kumawat
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1 Answers1

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$$\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{23x-11x^2}{(2x-1)(3-x)(3+x)}$$

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

Siong Thye Goh
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    oops, thanks for pointing that out. – Siong Thye Goh Apr 30 '19 at 14:48
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    I think many would appreciate a short justification for why this works at all. It seems like black magic, and people unfamiliar with the trick have no reason to trust it, nor any help remembering the details of it, nor any hope of figuring out what to do if we have a square factor in the denominator. – Arthur Apr 30 '19 at 14:52