How do you write the partial fraction decomposition of the rational expression # (5x+7)/(x^2+4x-5)#?

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Answer 1 · 2015-12-13T19:13:17

#(5x+7)/(x^2+4x-5)=3/(x+5)+2/(x-1)#

The denominator may be factorized as a trinomial as follows :

#x^2+4x-5=(x+5)(x-1)#.

Since these are 2 different linear factors, we may write the original expression in terms of partial fractions as follows :

#(5x+7)/(x^2+4x-5)=A/(x+5)+B/(x-1)#

#=(A(x-1)+B(x+5))/((x+5)(x-1))#

#=(Ax-A+Bx+5B)/((x+5)(x-1))#

#=((A+B)x+(5B-A))/((x+5)(x-1))#

Now comparing terms in the numerator yields the following set of linear equations which may be solved simultaneously :

#A+B=5 and 5B-A=7#

Solving we get :

#A=5-B=>5B-(5-B)=7=>B=12/6=2#.

#therefore A=5-2=3#.

Hence the partial fraction decomposition is

#(5x+7)/(x^2+4x-5)=3/(x+5)+2/(x-1)#