What is the partial-fraction decomposition of #(5x+7)/(x^2+4x-5)#?

1 Answer
Tony B
Oct 18, 2015

Part solution to start you on the way once you have seen the method. I take you up to #9/(x+5) +B/(x-1)#

Explanation:

Consider #x^2+4x-5#
This may be factorised into #(x+5)(x-1)#

Consequently we can write:

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

So:# (A(x-1) + B(x+5))/((x+5)(x-1)) =(5x + 7)/((x+5)(x-1))#

As the denominators on both sides of the equals are of the same value then so are the numerators. Consequently just considering the numerators we have:

# A(x-1) + B(x+5) =(5x + 7)#.................(1)

#Ax +Bx -A +5B =5x+7#

The #x# elements must equal each other and likewise the constant elements must also equal each other. So we have:

#Ax + Bx = 5x#.............................. (2)
#5B-A=7#.........................................(3)

Find the value of B from (3) and substitute into (2). Then with a bit of algebraic manipulation you have:

#A=18/4 = 9/2#

Substituting this back into (1) gives:

#9(x-1) +B(x+5)=5x+7#..............(4)

I will let you work out the value of B to sub into

#9/(x+5) +B/(x-1)#

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