# How do you express 1/(x^4 - 16) in partial fractions?

Jun 16, 2016

Derive a system of linear equations and solve to find:

$\frac{1}{{x}^{4} - 16} = \frac{1}{32 \left(x - 2\right)} - \frac{1}{32 \left(x + 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$

#### Explanation:

Assuming that we want partial fractions with Real coefficients, start by finding the Real factors of ${x}^{4} - 16$.

Use the difference of squares identity, which can be written:

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

twice, as follows:

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

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

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

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

Since these factors are distinct, we are looking for a partial fraction decomposition of the form:

$\frac{1}{{x}^{4} - 16} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{C x + D}{{x}^{2} + 4}$

$= \frac{A \left(x + 2\right) \left({x}^{2} + 4\right) + B \left(x - 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left(x - 2\right) \left(x + 2\right)}{{x}^{4} - 16}$

$= \frac{A \left(x + 2\right) \left({x}^{2} + 4\right) + B \left(x - 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left({x}^{2} - 4\right)}{{x}^{4} - 16}$

$= \frac{\left(A + B + C\right) {x}^{3} + \left(2 A - 2 B + D\right) {x}^{2} + \left(4 A + 4 B - 4 C\right) x + \left(8 A + 8 B - 4 D\right)}{{x}^{4} - 16}$

Equating coefficients, we get the following system of linear equations:

$\left\{\begin{matrix}A + B + C = 0 \\ 2 A - 2 B + D = 0 \\ 4 A + 4 B - 4 C = 0 \\ 8 A + 8 B - 4 D = 1\end{matrix}\right.$

We can write this system of equations as an augmented matrix, then solve it by performing a sequence of row operations to make the left hand side into a $4 \times 4$ identity matrix. Then the right hand column will be $\left(\begin{matrix}A \\ B \\ C \\ D\end{matrix}\right)$

Apologies for the length, but Gaussian elimination is like that...

$\left(\begin{matrix}1 & 1 & 1 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 4 & 4 & - 4 & 0 & | & 0 \\ 8 & - 8 & 0 & - 4 & | & 1\end{matrix}\right)$

Subtract $4 \times$row 1 from row 3 to get:

$\left(\begin{matrix}1 & 1 & 1 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 0 & 0 & - 8 & 0 & | & 0 \\ 8 & - 8 & 0 & - 4 & | & 1\end{matrix}\right)$

Divide row 3 by $- 8$ to get:

$\left(\begin{matrix}1 & 1 & 1 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 8 & - 8 & 0 & - 4 & | & 1\end{matrix}\right)$

Subtract row 3 from row 1 to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 8 & - 8 & 0 & - 4 & | & 1\end{matrix}\right)$

Subtract $4 \times$row 2 from row 4 to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & - 8 & | & 1\end{matrix}\right)$

Divide row 4 by $- 8$ to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 2 & - 2 & 0 & 1 & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & - \frac{1}{8}\end{matrix}\right)$

Subtract $2 \times$ row 1 from row 2 to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 0 & - 4 & 0 & 1 & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & - \frac{1}{8}\end{matrix}\right)$

Divide row 2 by $- 4$ to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & - \frac{1}{4} & | & 0 \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & - \frac{1}{8}\end{matrix}\right)$

Add $\frac{1}{4} \times$row 4 to row 2 to get:

$\left(\begin{matrix}1 & 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & 0 & | & - \frac{1}{32} \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & - \frac{1}{8}\end{matrix}\right)$

Subtract row 2 from row 1 to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & | & \frac{1}{32} \\ 0 & 1 & 0 & 0 & | & - \frac{1}{32} \\ 0 & 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & - \frac{1}{8}\end{matrix}\right)$

So we find:

$\left(\begin{matrix}A \\ B \\ C \\ D\end{matrix}\right) = \left(\begin{matrix}\frac{1}{32} \\ - \frac{1}{32} \\ 0 \\ - \frac{1}{8}\end{matrix}\right)$

Hence:

$\frac{1}{{x}^{4} - 16} = \frac{1}{32 \left(x - 2\right)} - \frac{1}{32 \left(x + 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$

Jun 17, 2016

$\frac{1}{{x}^{4} - 16} = \frac{1}{32 \left(x - 2\right)} - \frac{1}{32 \left(x + 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$

#### Explanation:

Use the difference of squares identity:

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

to factor the denominator.

To make things easier, break down into partial fractions in two stages:

$\frac{1}{{x}^{4} - 16} = \frac{1}{\left({x}^{2} - 4\right) \left({x}^{2} + 4\right)}$

We find:

$\frac{1}{{x}^{2} - 4} - \frac{1}{{x}^{2} + 4} = \frac{\left({x}^{2} + 4\right) - \left({x}^{2} - 4\right)}{{x}^{4} - 16} = \frac{8}{{x}^{4} - 16}$

So:

$\frac{1}{{x}^{4} - 16} = \frac{1}{8 \left({x}^{2} - 4\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$

Then note that:

$\frac{1}{x - 2} - \frac{1}{x + 2} = \frac{\left(x + 2\right) - \left(x - 2\right)}{{x}^{2} - 4} = \frac{4}{{x}^{2} - 4}$

So:

$\frac{1}{{x}^{2} - 4} = \frac{1}{4 \left(x - 2\right)} - \frac{1}{4 \left(x + 2\right)}$

And:

$\frac{1}{8 \left({x}^{2} - 4\right)} = \frac{1}{32 \left(x - 2\right)} - \frac{1}{32 \left(x + 2\right)}$

Putting it all together:

$\frac{1}{{x}^{4} - 16} = \frac{1}{32 \left(x - 2\right)} - \frac{1}{32 \left(x + 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$