How do you evaluate #\frac { ( x + 1) ^ { 2} } { ( x - 2) ( x - 1) ^ { 2} } = \frac { A } { x - 2} + \frac {B}{x-1}+\frac { c } { ( x - 1) ^ { 2} }#?

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Answer 1 · 2017-08-01T15:25:55

The answers are #A=9, #B=-8# and #C=-4#

#(x+1)^2/((x-2)(x-1)^2)=(x^2+2x+1)/((x-2)(x-1)^2)#

#=A/(x-2)+B/(x-1)+C/(x-1)^2#

#=(A(x-1)^2+B(x-2)(x-1)+C(x-2))/((x-2)(x-1)^2)#

The denominators are the same, we compare the numerators

#x^2+2x+1=A(x-1)^2+B(x-2)(x-1)+C(x-2)#

Let #x=1#, #=>#, #4=-C#, #=>#, #C=-4#

Let #x=2#, #=>#, #9=A#

Coefficient of #x^2#

#1=A+B#, #=>#, #B=1-A=1-9=-8#

Therefore,

#(x+1)^2/((x-2)(x-1)^2)=9/(x-2)-8/(x-1)-4/(x-1)^2#