# How do you write the partial fraction decomposition of the rational expression (x² + 2x-7) /( (x+3)(x-1)²)?

Feb 28, 2017

The answer is $= \frac{- \frac{1}{4}}{x + 3} - \frac{1}{x - 1} ^ 2 + \frac{\frac{5}{4}}{x - 1}$

#### Explanation:

Let 's perform the decomposition into partial fractions

$\frac{{x}^{2} + 2 x - 7}{\left(x + 3\right) {\left(x - 1\right)}^{2}} = \frac{A}{x + 3} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$

$= \frac{A {\left(x - 1\right)}^{2} + B \left(x + 3\right) + C \left(x + 3\right) \left(x - 1\right)}{\left(x + 3\right) {\left(x - 1\right)}^{2}}$

As the denominators are the same, we compare the numerators

${x}^{2} + 2 x - 7 = A {\left(x - 1\right)}^{2} + B \left(x + 3\right) + C \left(x + 3\right) \left(x - 1\right)$

Let $x = - 3$, $\implies$, $- 4 = 16 A$, $\implies$, $A = - \frac{1}{4}$

Let $x = 1$, $\implies$, $- 4 = 4 B$, $\implies$, $B = - 1$

Coefficients of ${x}^{2}$

$1 = A + C$, $\implies$, $C = 1 - A = 1 + \frac{1}{4} = \frac{5}{4}$

Therefore,

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