How do you write the partial fraction decomposition of the rational expression # (4x+23)/(x^2-x-6)#?

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Answer 1 · 2015-12-25T04:05:16

#(4x+23)/(x^2-x-6)=7/(x-3)-3/(x+2)#

Factor the denominator and split apart the expression.

#(4x+23)/((x-3)(x+2))=A/(x-3)+B/(x+2)#

Find a common denominator of #(x-3)(x+2)#.

#4x+23=A(x+2)+B(x-3)#

Plug in #-2# for #x#:

#4(-2)+23=B(-2-3)#
#B=-3#

Plug in #3# for #x#:

#4(3)+23=A(3+2)#
#A=7#

Plug this into the original expression:

#(4x+23)/(x^2-x-6)=7/(x-3)-3/(x+2)#