# How do you write the partial fraction decomposition of the rational expression  6/(x^2-25)^2?

Dec 30, 2015

$\frac{6}{{x}^{2} - 25} ^ 2$

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

#### Explanation:

Note that:

${\left({x}^{2} - 25\right)}^{2} = {\left(\left(x - 5\right) \left(x + 5\right)\right)}^{2}$

So need to solve:

$\frac{6}{{x}^{2} - 25} ^ 2 = \frac{A}{x - 5} + \frac{B}{x - 5} ^ 2 + \frac{C}{x + 5} + \frac{D}{x + 5} ^ 2$

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

$= \frac{\left(A x + \left(B - 5 A\right)\right) {\left(x + 5\right)}^{2} + \left(C x + \left(D + 5 C\right)\right) {\left(x - 5\right)}^{2}}{{x}^{2} - 25} ^ 2$

$= \frac{\left(A x + \left(B - 5 A\right)\right) \left({x}^{2} + 10 x + 25\right) + \left(C x + \left(D + 5 C\right)\right) \left({x}^{2} - 10 x + 25\right)}{{x}^{2} - 25} ^ 2$

$= \frac{\left(A + C\right) {x}^{3} + \left(B + D + 5 A - 5 C\right) {x}^{2} + \left(10 B - 10 D - 25 A - 25 C\right) x + 25 \left(B + D - 5 A + 5 C\right)}{{x}^{2} - 25} ^ 2$

Hence:

$A + C = 0$

$B + D + 5 A - 5 C = 0$

$10 B - 10 D - 25 A - 25 C = 0$

$25 \left(B + D - 5 A + 5 C\right) = 6$

From the first of these $C = - A$, so the rest become:

$B + D + 10 A = 0$

$10 \left(B - D\right) = 0$

$25 \left(B + D - 10 A\right) = 6$

From the second of these $D = B$, so the rest become:

$2 B + 10 A = 0$

$25 \left(2 B - 10 A\right) = 6$

From the first of these $B = - 5 A$, so the second becomes:

$25 \left(- 20 A\right) = 6$

Hence $A = - \frac{3}{250}$, $B = \frac{3}{50}$, $C = \frac{3}{250}$, $D = \frac{3}{50}$

So:

$\frac{6}{{x}^{2} - 25} ^ 2$

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