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

Oct 18, 2015

Part solution to start you on the way once you have seen the method. I take you up to $\frac{9}{x + 5} + \frac{B}{x - 1}$

#### Explanation:

Consider ${x}^{2} + 4 x - 5$
This may be factorised into $\left(x + 5\right) \left(x - 1\right)$

Consequently we can write:

$\frac{A}{x + 5} + \frac{B}{x - 1} = \frac{5 x + 7}{\left(x + 5\right) \left(x - 1\right)} = \frac{5 x + 7}{{x}^{2} + 4 x - 5}$

So:$\frac{A \left(x - 1\right) + B \left(x + 5\right)}{\left(x + 5\right) \left(x - 1\right)} = \frac{5 x + 7}{\left(x + 5\right) \left(x - 1\right)}$

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:

$A \left(x - 1\right) + B \left(x + 5\right) = \left(5 x + 7\right)$.................(1)

$A x + B x - A + 5 B = 5 x + 7$

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

$A x + B x = 5 x$.............................. (2)
$5 B - A = 7$.........................................(3)

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

$A = \frac{18}{4} = \frac{9}{2}$

Substituting this back into (1) gives:

$9 \left(x - 1\right) + B \left(x + 5\right) = 5 x + 7$..............(4)

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

$\frac{9}{x + 5} + \frac{B}{x - 1}$