# How do you write the partial fraction decomposition of the rational expression  (x+4)/((x+1)(x-2)^2 )?

Feb 12, 2016

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

#### Explanation:

firstly note that  x-2)^2
has factors (x-2) and ${\left(x - 2\right)}^{2}$

the factors of the denominator are linear hence the numerators are constants.

$\Rightarrow \frac{x + 4}{\left(x + 1\right) {\left(x - 2\right)}^{2}} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x - 2} ^ 2$

now multiply through by $\left(x + 1\right) {\left(x - 2\right)}^{2}$

hence : x+4 = $A {\left(x - 2\right)}^{2} + B \left(x + 1\right) \left(x - 2\right) + C \left(x + 1\right) \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

the aim now is to find values for A , B and C . Note that if x = 2 then the terms with A and B will be zero , and if x = -1 the terms with B and C will be zero. This is the starting point in finding values.

let x = 2 in (1) : $6 = 3 C \Rightarrow C = 2$

let x = -1 in (1) : -3 =$9 A \Rightarrow A = \frac{1}{3}$
any value for x may be chosen to find B.

let x = 0 in (1) : 4 = 4A - 2B + C $\Rightarrow 2 B = 4 A + C - 4$

hence 2B $= \frac{4}{3} + 2 - 4 = - \frac{2}{3} \Rightarrow B = - \frac{1}{3}$

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