Alternative efficient ways to do partial fraction?

[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?

 

[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}}<br /> {{x^3 - 7x + 3}}

So you start by factoring and introducing the unknown coëfficiënts:

\frac{{x + 1}}<br /> {{x^3 - 7x + 3}} = \frac{{x + 1}}<br /> {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{A}<br /> {{x - 1}} + \frac{B}<br /> {{x - 2}} + \frac{C}<br /> {{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)}}<br /> {{\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}<br /> x = 1 \Rightarrow A\left( {1 - 2} \right)\left( {1 + 3} \right) = 2 \Leftrightarrow A = - \frac{1}<br /> {2} \hfill \\<br /> x = 2 \Rightarrow B\left( {2 - 1} \right)\left( {2 + 3} \right) = 3 \Leftrightarrow B = \frac{3}<br /> {5} \hfill \\<br /> x = - 3 \Rightarrow C\left( { - 3 - 1} \right)\left( { - 3 - 2} \right) = - 2 \Leftrightarrow C = - \frac{1}<br /> {{10}} \hfill \\ <br /> \end{gathered}

Which gives the result:

\frac{{x + 1}}<br /> {{x^3 - 7x + 3}} = \frac{{ - 1}}<br /> {{2\left( {x - 1} \right)}} + \frac{3}<br /> {{5\left( {x - 2} \right)}} - \frac{1}<br /> {{10\left( {x + 3} \right)}}