# How do you express (x+4)/((x+1)(x-2)^2) in partial fractions?

May 8, 2018

The answer is $= \frac{\frac{13}{3}}{x + 1} + \frac{42}{x - 2} ^ 2 + \frac{- \frac{13}{3}}{x - 2}$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x + 40}{\left(x + 1\right) {\left(x - 2\right)}^{2}} = \frac{A}{x + 1} + \frac{B}{x - 2} ^ 2 + \frac{C}{x - 2}$

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

The denominators are the same, compare the numerators

$x + 40 = A {\left(x - 2\right)}^{2} + B \left(x + 1\right) + C \left(x + 1\right) \left(x - 2\right)$

Let $x = - 1$, $\implies$, $39 = 9 A$, $\implies$, $A = \frac{39}{9} = \frac{13}{3}$

Let $x = 2$, $\implies$, $42 = 3 B$, $\implies$, $B = \frac{42}{3} = 14$

Coefficients of ${x}^{2}$

$0 = A + C$, $\implies$, $C = - A = - \frac{13}{3}$

Finally,

$\frac{x + 40}{\left(x + 1\right) {\left(x - 2\right)}^{2}} = \frac{\frac{13}{3}}{x + 1} + \frac{42}{x - 2} ^ 2 + \frac{- \frac{13}{3}}{x - 2}$