How do you find the partial fraction decomposition when you have repeated quadratic or linear factors?
1 Answer
Imagine the partial fraction decomposition problem:
(2x-1)/(x^2-x-6)2x−1x2−x−6
Here, the denominator would simplify into
(2x-1)/((x-3)(x+2))=A/(x-3)+B/(x-2)2x−1(x−3)(x+2)=Ax−3+Bx−2
However, when the denominator has a repeated factor, something slightly different happens.
Consider
(x+3)/(x^2+4x+4)x+3x2+4x+4
Since
(x+3)/((x+2)(x+2))=A/(x+2)+B/(x+2)x+3(x+2)(x+2)=Ax+2+Bx+2
However, this will not work. This is remedied by "counting up" to whatever the exponent is on the repeated factor. This would look like
(x+3)/(x+2)^2=A/(x+2)+B/(x+2)^2x+3(x+2)2=Ax+2+B(x+2)2
Which can indeed be solved.
An example with a third power:
(3x-2)/(2x-1)^3=A/(2x-1)+B/(2x-1)^2+C/(2x-1)^33x−2(2x−1)3=A2x−1+B(2x−1)2+C(2x−1)3
This is very similar for irreducible quadratic factors, except for that they take the form
For example:
(5x)/((x^2+4)^2(x-3))=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2+E/(x-3)5x(x2+4)2(x−3)=Ax+Bx2+4+Cx+D(x2+4)2+Ex−3