Question

How do you express `(x^2)/((x^4) - (2x^2) - 8)` in partial fractions?

Answer 1

`x^2/(x^4 - 2x^2 - 8) = 1/(6(x - 2)) - 1/(6(x+ 2)) + 1/(3(x^2 + 2))`

Explanation:

If we let `u = x^2`, then we can rewrite as

`u/(u^2 - 2u - 8)`

The denominator can be factored as `(u - 4)(u + 2)`. Hence:

`A/(u - 4) + B/(u + 2) = u/(u^2 - 2u - 8)`

`A(u + 2) + B(u - 4) = u`

`Au + 2A + Bu - 4B = u`

We can now write a system of equations:

`{(A + B = 1), (2A - 4B = 0):}`

Simplify the second equation to get:

`{(A + B = 1), (A - 2B = 0):}`

Substitute `A = 1- B` into the second equation.

`1 - B - 2B = 0`

`-3B = -1`

`B = 1/3`

This means that `A = 2/3`.

Therefore, we can say

`u/(u^2 - 2u - 8) = 2/(3(u - 4)) + 1/(3(u + 2))`

Reversing our initial substitution:

`x^2/(x^4 - 2x^2 - 8) =2/(3(x^2 - 4)) + 1/(3(x^2 + 2))`

But we now see that `x^2 - 4 = (x+ 2)(x - 2)`, so we can simplify further.

`A/(x +2) + B/(x- 2) = (2/3)/((x +2)(x - 2))`

`A(x - 2) + B(x + 2) = 2/3`

`Ax - 2A + Bx + 2B = 2/3`

Then we have a new system of equations:

`{(A + B = 0), (B - A= 1/3):}`

From the first equation we deduce `B = -A` which means that `-2A = 1/3` and that `A = -1/6`. Accordingly, `B = 1/6`.

Hence:

`2/(3(x^2 - 4)) = 1/(6(x- 2)) - 1/(6(x+ 2))`

Putting everything back together, we get:

`x^2/(x^4 - 2x^2 - 8) = 1/(6(x - 2)) - 1/(6(x+ 2)) + 1/(3(x^2 + 2))`

Hopefully this helps!