How do you write the partial fraction decomposition of the rational expression (2x^2-18x-12)/(x^3-4x)?

1 Answer
George C.
Aug 7, 2016

(2x^2-18x-12)/(x^3-4x)=3/x-5/(x-2)+4/(x+2)

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)

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