How do you decompose #(x-6)/(x^2-2x)# into partial fractions?

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Answer 1 · 2016-12-21T22:06:40

Begin with this equation:

#(x - 6)/(x^2 - 2x) = A/x + B/(x - 2)#

Multiply both sides of the equation by #x(x - 2)#:

#x - 6 = A(x - 2) + Bx#

Let x = 0:

#0 - 6 = A(0 - 2) + B(0)#

#-6 = A(-2)#

#A = 3#

Let x = 2:

#2 - 6 = A(2 - 2) + B(2)#

#-4 = 2B#

#B = -2#

#A = 3 and B = -2#

Check:

#3/x - 2/(x - 2) =#

#3/x(x - 2)/(x - 2) - 2/(x - 2)x/x =#

#(3x - 6)/(x(x - 2)) - (2x)/(x(x - 2)) =#

#(3x - 6 - 2x)/(x(x - 2)) =#

#(x - 6)/(x^2 - 2x)#

The checks.

The answer is:

#(x - 6)/(x^2 - 2x) = 3/x - 2/(x - 2)#