What is the partial-fraction decomposition of (5x+7)/(x^2+4x-5)5x+7x2+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)9x+5+Bx−1

Explanation:

Consider x^2+4x-5x2+4x−5
This may be factorised into (x+5)(x-1)(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)Ax+5+Bx−1=5x+7(x+5)(x−1)=5x+7x2+4x−5

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

Ax +Bx -A +5B =5x+7Ax+Bx−A+5B=5x+7

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

Ax + Bx = 5xAx+Bx=5x.............................. (2)
5B-A=75B−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 = 9/2A=184=92

Substituting this back into (1) gives:

9(x-1) +B(x+5)=5x+79(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)9x+5+Bx−1

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