# How do you write the partial fraction decomposition of the rational expression  (x^2 -111)/ (x^4- x^2- 72)?

Dec 15, 2015

20/(17(x-3))-20/(17(x+3))-103/(17(x^2+8)

#### Explanation:

Factor the denominator.

${x}^{4} - {x}^{2} - 72 = \left({x}^{2} - 9\right) \left({x}^{2} + 8\right) = \left(x + 3\right) \left(x - 3\right) \left({x}^{2} + 8\right)$

Write the partial fraction decomposition expression.

$\frac{{x}^{2} - 111}{\left(x + 3\right) \left(x - 3\right) \left({x}^{2} + 8\right)} = \frac{A}{x + 3} + \frac{B}{x - 3} + \frac{C x + D}{{x}^{2} + 8}$

${x}^{2} - 111 = A \left(x - 3\right) \left({x}^{2} + 8\right) + B \left(x + 3\right) \left({x}^{2} + 8\right) + \left(C x + D\right) \left({x}^{2} - 9\right)$

${x}^{2} - 111 = A \left({x}^{3} - 3 {x}^{2} + 8 x - 24\right) + B \left({x}^{3} + 3 {x}^{2} + 8 x + 24\right) + C {x}^{3} - 9 C x + D {x}^{2} - 9 D$

${x}^{2} - 111 = {x}^{3} \left(A + B + C\right) + {x}^{2} \left(- 3 A + 3 B + D\right) + x \left(8 A + 8 B - 9 C\right) + 1 \left(- 24 A + 24 B - 9 D\right)$

The following system can be deduced:

$\left\{\begin{matrix}A + B + C = 0 \\ - 3 A + 3 B + D = 1 \\ 8 A + 8 B - 9 C = 0 \\ - 24 A + 24 B - 9 D = 111\end{matrix}\right.$

Solve the system:

$\left\{\begin{matrix}A = - \frac{20}{17} \\ B = \frac{20}{17} \\ C = 0 \\ D = - \frac{103}{17}\end{matrix}\right.$

Plug in these values:

 (x^2 -111)/ (x^4- x^2- 72)=20/(17(x-3))-20/(17(x+3))-103/(17(x^2+8)