# Alternative efficient ways to do partial fraction?

1. Sep 28, 2005

### hanhao

what are alternative ways to do partial fraction?
other than the normal method of mulplying the whole term over and comparing coefficient which is super headache method

i heard of "cover-up method", gets the answer real quick

anyone have details to this?

2. Sep 28, 2005

### TD

By choosing "smart zeroes" of the denominator, you can speed up the process but this won't always work (fully).

For example, we wish to expand:

$$\frac{{x + 1}} {{x^3 - 7x + 3}}$$

So you start by factoring and introducing the unknown coëfficiënts:

$$\frac{{x + 1}} {{x^3 - 7x + 3}} = \frac{{x + 1}} {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{A} {{x - 1}} + \frac{B} {{x - 2}} + \frac{C} {{x + 3}} = \frac{{A\left( {x - 2} \right)\left( {x + 3} \right) + B\left( {x - 1} \right)\left( {x + 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right)}} {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}}$$

Then you set up the equation for the numerators to be the same:

$$A\left( {x - 2} \right)\left( {x + 3} \right) + B\left( {x - 1} \right)\left( {x + 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right) = x + 1$$

Traditionally, you would now expand the entire left side and then group in powers of x to get a system, which is still doable.

Another way to go on now would be to choose some values for x, preferably values which may simplify things, i.e. zeroes of the denominator:

$$\begin{gathered} x = 1 \Rightarrow A\left( {1 - 2} \right)\left( {1 + 3} \right) = 2 \Leftrightarrow A = - \frac{1} {2} \hfill \\ x = 2 \Rightarrow B\left( {2 - 1} \right)\left( {2 + 3} \right) = 3 \Leftrightarrow B = \frac{3} {5} \hfill \\ x = - 3 \Rightarrow C\left( { - 3 - 1} \right)\left( { - 3 - 2} \right) = - 2 \Leftrightarrow C = - \frac{1} {{10}} \hfill \\ \end{gathered}$$

Which gives the result:

$$\frac{{x + 1}} {{x^3 - 7x + 3}} = \frac{{ - 1}} {{2\left( {x - 1} \right)}} + \frac{3} {{5\left( {x - 2} \right)}} - \frac{1} {{10\left( {x + 3} \right)}}$$