Question

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

Answer 1

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)`