# How do you write the partial fraction decomposition of the rational expression (3x^2+2x)/(x^2-4)?

Nov 26, 2016

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

#### Explanation:

As the degree of the numerator is $=$ the degree of the denominator

Let's do a long division

$\textcolor{w h i t e}{a a a a a}$$3 {x}^{2} + 2 x$$\textcolor{w h i t e}{a a a a a}$∣${x}^{2} - 4$

$\textcolor{w h i t e}{a a a a a}$$3 {x}^{2} - 12$$\textcolor{w h i t e}{a a a a a}$∣$3$

$\textcolor{w h i t e}{a a a a a a}$$0 + 2 x + 12$

$\frac{3 {x}^{2} + 2 x}{{x}^{2} - 4} = 3 + \frac{2 x + 12}{{x}^{2} - 4}$

Let's factorise the denominator

x^2-4=(x+2)(x-2)#

Let's do the decomposition in partial fractions

$\frac{2 x + 12}{{x}^{2} - 4} = \frac{2 x + 12}{\left(x + 2\right) \left(x - 2\right)} = \frac{A}{x + 2} + \frac{B}{x - 2}$

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

So, $\left(2 x + 12\right) = \left(A \left(x - 2\right) + B \left(x + 2\right)\right)$

Let $x = 2$, $16 = 4 B$, $\implies$, $B = 4$

Let $x = - 2$, $8 = - 4 A$, $\implies$, $A = - 2$

So, $\frac{2 x + 12}{{x}^{2} - 4} = - \frac{2}{x + 2} + \frac{4}{x - 2}$

And finally we have

$\frac{3 {x}^{2} + 2 x}{{x}^{2} - 4} = 3 - \frac{2}{x + 2} + \frac{4}{x - 2}$