# How do you express 6/(x^2-25)^2 in partial fractions?

Oct 28, 2016

$\frac{6}{{x}^{2} - 25} ^ 2 = \frac{3}{50 {\left(x + 5\right)}^{2}} + \frac{3}{250 \left(x + 5\right)} + \frac{3}{50 {\left(x - 5\right)}^{2}} - \frac{3}{250 \left(x - 5\right)}$

#### Explanation:

we start by factorising the denominator
${\left({x}^{2} - 25\right)}^{2} = {\left(\left(x + 5\right) \left(x - 5\right)\right)}^{2}$
So we have$= 6 \left(\frac{1}{{\left(x + 5\right)}^{2} {\left(x - 5\right)}^{2}}\right)$
Then we can do the decomposition in partial fractions
$\frac{1}{{\left(x + 5\right)}^{2} {\left(x - 5\right)}^{2}} = \frac{A}{{\left(x + 5\right)}^{2}} + \frac{B}{x + 5} + \frac{C}{{\left(x - 5\right)}^{2}} + \frac{D}{x - 5}$

$= \frac{\left(A {\left(x - 5\right)}^{2} + B \left(x + 5\right) {\left(x - 5\right)}^{2} + C {\left(x - 5\right)}^{2} + D \left(x - 5\right) {\left(x + 5\right)}^{2}\right)}{\left({\left(x + 5\right)}^{2} {\left(x - 5\right)}^{2}\right)}$

So 1=(A(x-5)^2+B(x+5)(x-5)^2+C(x+5)^2+D(x-5)(x+5)^2))
Let $x = 5$ we get $1 = 100 C$ $\implies$ $C = \frac{1}{100}$
Let $x = - 5$ then $1 = 100 A$ $\implies$ $A = \frac{1}{100}$
Let $x = 0$ then $1 = 25 A + 125 B + 25 C - 125 D$
$1 = \frac{25}{100} + \frac{25}{100} + 125 \left(B - D\right)$
$B - D = \frac{1}{250}$
Coefficients of ${x}^{3}$ $\implies$ $0 = B + D$$\implies$$B = - D$
$B = \frac{1}{500}$ and $D = - \frac{1}{500}$
So the final result is $= \frac{6}{100 {\left(x + 5\right)}^{2}} + \frac{6}{500 \left(x + 5\right)} + \frac{6}{100 {\left(x - 5\right)}^{2}} - \frac{6}{500 \left(x - 5\right)}$