How do you express #(2x^2-18x-12)/(x^3-4x)# in partial fractions?

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Answer 1 · 2016-10-30T06:43:37

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

Expand:
#(2x^2 - 18x - 12)/(x^3 - 4x) = A/x + B/(x + 2) + C/(x - 2)#

Multiply both sides by #x(x + 2)(x - 2)#

#2x^2 - 18x - 12 = A(x + 2)(x - 2) + B(x)(x - 2) + C(x)(x + 2)#

Let x = 0 to make B and C disappear:

#-12 = A(2)(-2)#

#A = 3#

Let x = -2 to make A and C disappear:

#2(-2)^2 - 18(-2) - 12 = B(-2)(-2 - 2)#

#32 = B(8)#

#B = 4#

Let x = 2 to make A and B disappear:

#2(2)^2 - 18(2) - 12 = C(2)(2 + 2)#

#-40 = C(8)#

#C = -5#

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