How do you express  (6x^2+7x-6)/((x^2-4)(x+2)) in partial fractions?

Feb 14, 2016

$\frac{2}{x - 2} + \frac{4}{x + 2} - \frac{1}{x + 2} ^ 2$

Explanation:

first step here is to factor the denominator

$\left({x}^{2} - 4\right) \left(x + 2\right) = \left(x - 2\right) \left(x + 2\right) \left(x + 2\right) = \left(x - 2\right) {\left(x + 2\right)}^{2}$
Note also that the factors of ${\left(x + 2\right)}^{2} = \left(x + 2\right) \textcolor{b l a c k}{\text{ and}} {\left(x + 2\right)}^{2}$
Since these factors are linear , the numerators will be constants , say A , B and C.

$\frac{6 {x}^{2} + 7 x - 6}{\left(x - 2\right) {\left(x + 2\right)}^{2}} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{C}{x + 2} ^ 2$

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

the following equation will be referred to as ( 1)
$6 {x}^{2} + 7 x - 6 = A {\left(x + 2\right)}^{2} + B \left(x - 2\right) \left(x + 2\right) + C \left(x - 2\right)$

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

let x = 2 in (1) : 32 = 16A $\Rightarrow A = 4$

let x = -2 in (1) : 4 = - 4C $\Rightarrow C = - 1$

Any value for x , may be chosen to find B.

let x = 0 in (1) : - 6 = 4A - 4B - 2C

hence 4B = 4A - 2C + 6 = 8 + 2 + 6 = 16 $\Rightarrow B = 4$

$\Rightarrow \frac{6 {x}^{2} + 7 x - 6}{\left({x}^{2} - 4\right) \left(x + 2\right)} = \frac{2}{x - 2} + \frac{4}{x + 2} - \frac{1}{x + 2} ^ 2$