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