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

Dec 13, 2015

$\frac{5 x + 7}{{x}^{2} + 4 x - 5} = \frac{3}{x + 5} + \frac{2}{x - 1}$

#### Explanation:

The denominator may be factorized as a trinomial as follows :

${x}^{2} + 4 x - 5 = \left(x + 5\right) \left(x - 1\right)$.

Since these are 2 different linear factors, we may write the original expression in terms of partial fractions as follows :

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

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

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

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

Now comparing terms in the numerator yields the following set of linear equations which may be solved simultaneously :

$A + B = 5 \mathmr{and} 5 B - A = 7$

Solving we get :

$A = 5 - B \implies 5 B - \left(5 - B\right) = 7 \implies B = \frac{12}{6} = 2$.

$\therefore A = 5 - 2 = 3$.

Hence the partial fraction decomposition is

$\frac{5 x + 7}{{x}^{2} + 4 x - 5} = \frac{3}{x + 5} + \frac{2}{x - 1}$