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

Feb 11, 2017

The answer is $= \frac{\frac{19}{4}}{x - 3} ^ 2 + \frac{\frac{9}{16}}{x - 3} + \frac{- \frac{9}{16}}{x + 1}$

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x = 3$, $\implies$, $19 = 4 A$, $\implies$, $A = \frac{19}{4}$

Let $x = - 1$, $\implies$, $- 9 = 16 C$, $\implies$, $C = - \frac{9}{16}$

Coefficients of ${x}^{2}$

$0 = B + C$, $\implies$, $B = - C = \frac{9}{16}$

Therefore,

$\frac{7 x - 2}{{\left(x - 3\right)}^{2} \left(x + 1\right)} = \frac{\frac{19}{4}}{x - 3} ^ 2 + \frac{\frac{9}{16}}{x - 3} + \frac{- \frac{9}{16}}{x + 1}$