How do you express #(x^2 - 8x + 44) / ((x + 2) (x - 2)^2)# in partial fractions?

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Answer 1 · 2018-01-02T20:31:33

#(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2#

#(x^2-8x+44)/[(x+2)*(x-2)^2]#

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

After expanding denominator,

#A*(x-2)^2+B*(x^2-4)+C*(x+2)=x^2-8x+44#

Set #x=-2#, #16A=64#, so #A=4#

Set #x=2#, #4C=32#, so #C=8#

Set #x=0#, #4A-4B+2C=44#, so #B=-3#

Thus,

#(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2#