# How do you write the partial fraction decomposition of the rational expression  (3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)?

Dec 3, 2016

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

#### Explanation:

To factorise the denominator,

We use ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

and ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

So,

${x}^{4} - 8 {x}^{2} + 16 = {\left({x}^{2} - 4\right)}^{2} = {\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}$

Therefore,

$\frac{3 {x}^{2} + 12 x - 20}{{x}^{4} - 8 {x}^{2} + 16} = \frac{3 {x}^{2} + 12 x - 20}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

We can now do the decomposition in partial fractions

$\frac{3 {x}^{2} + 12 x - 20}{{x}^{4} - 8 {x}^{2} + 16} = \frac{A}{x + 2} ^ 2 + \frac{B}{x + 2} + \frac{C}{x - 2} ^ 2 + \frac{D}{x - 2}$

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

Therefore,

$\left(3 {x}^{2} + 12 x - 20\right) = A {\left(x - 2\right)}^{2} + B {\left(x - 2\right)}^{2} \left(x + 2\right) + C {\left(x + 2\right)}^{2} + D {\left(x + 2\right)}^{2} \left(x - 2\right)$

Let $x = 2$, $\implies$, $16 = 16 C$, $\implies$, $C = 1$

Let $x = - 2$, $\implies$, $- 32 = 16 A$, $\implies$, $A = - 2$

Let $x = 0$, $\implies$, $- 20 = 4 A + 8 B + 4 C - 8 D$

$B - D = - 2$

Coefficients of ${x}^{3}$

$0 = B + D$

So, $B = - 1$ and $D = 1$

So,

$\frac{3 {x}^{2} + 12 x - 20}{{x}^{4} - 8 {x}^{2} + 16} = - \frac{2}{x + 2} ^ 2 - \frac{1}{x + 2} + \frac{1}{x - 2} ^ 2 + \frac{1}{x - 2}$