# How do you express (x² + 2x-7 )/( (x+3)(x-1)²) in partial fractions?

Nov 29, 2016

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

#### Explanation:

Let's do the decomposition in partial fractions

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

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

Therefore,

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

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

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

Coefficients of ${x}^{2}$, $\implies$, $1 = A + B$

So, $B = 1 + \frac{1}{4} = \frac{5}{4}$

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