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

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

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