How do you write the partial fraction decomposition of the rational expression (2x^2-18x-12)/(x^3-4x)2x2−18x−12x3−4x?
1 Answer
Aug 7, 2016
Explanation:
(2x^2-18x-12)/(x^3-4x)2x2−18x−12x3−4x
=(2x^2-18x-12)/(x(x-2)(x+2))=2x2−18x−12x(x−2)(x+2)
=A/x+B/(x-2)+C/(x+2)=Ax+Bx−2+Cx+2
Use Heaviside's cover-up method to find:
A=(2(0)^2-18(0)-12)/(((0)-2)((0)+2)) = (-12)/(-4) = 3A=2(0)2−18(0)−12((0)−2)((0)+2)=−12−4=3
B=(2(2)^2-18(2)-12)/((2)((2)+2)) = (8-36-12)/8 = (-40)/8 = -5B=2(2)2−18(2)−12(2)((2)+2)=8−36−128=−408=−5
C=(2(-2)^2-18(-2)-12)/((-2)((-2)-2)) = (8+36-12)/8 = 32/8 = 4C=2(−2)2−18(−2)−12(−2)((−2)−2)=8+36−128=328=4
So:
(2x^2-18x-12)/(x^3-4x)=3/x-5/(x-2)+4/(x+2)2x2−18x−12x3−4x=3x−5x−2+4x+2
