Question

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

Answer 1

Derive a system of linear equations and solve to find:

`1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))`

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=(a-b)(a+b)`

twice, as follows:

`x^4-16 = (x^2)^2-4^2`

`= (x^2-4)(x^2+4)`

`= (x^2-2^2)(x^2+4)`

`= (x-2)(x+2)(x^2+4)`

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

`1/(x^4-16) = A/(x-2)+B/(x+2)+(Cx+D)/(x^2+4)`

`=(A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2))/(x^4-16)`

`=(A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x^2-4))/(x^4-16)`

`=((A+B+C)x^3+(2A-2B+D)x^2+(4A+4B-4C)x+(8A+8B-4D))/(x^4-16)`

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

`{ (A+B+C=0), (2A-2B+D=0), (4A+4B-4C=0), (8A+8B-4D=1) :}`

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 `4xx4` identity matrix. Then the right hand column will be `((A),(B),(C),(D))`

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

`((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (4, 4, -4, 0, |, 0), (8, -8, 0, -4, |, 1))`

Subtract `4xx`row 1 from row 3 to get:

`((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, -8, 0, |, 0), (8, -8, 0, -4, |, 1))`

Divide row 3 by `-8` to get:

`((1, 1, 1, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (8, -8, 0, -4, |, 1))`

Subtract row 3 from row 1 to get:

`((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (8, -8, 0, -4, |, 1))`

Subtract `4xx`row 2 from row 4 to get:

`((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, -8, |, 1))`

Divide row 4 by `-8` to get:

`((1, 1, 0, 0, |, 0), (2, -2, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))`

Subtract `2xx` row 1 from row 2 to get:

`((1, 1, 0, 0, |, 0), (0, -4, 0, 1, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))`

Divide row 2 by `-4` to get:

`((1, 1, 0, 0, |, 0), (0, 1, 0, -1/4, |, 0), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))`

Add `1/4xx`row 4 to row 2 to get:

`((1, 1, 0, 0, |, 0), (0, 1, 0, 0, |, -1/32), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))`

Subtract row 2 from row 1 to get:

`((1, 0, 0, 0, |, 1/32), (0, 1, 0, 0, |, -1/32), (0, 0, 1, 0, |, 0), (0, 0, 0, 1, |, -1/8))`

So we find:

`((A),(B),(C),(D)) = ((1/32),(-1/32),(0),(-1/8))`

Hence:

`1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))`

Answer 2

`1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))`

Explanation:

Use the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

to factor the denominator.

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

`1/(x^4-16) = 1/((x^2-4)(x^2+4))`

We find:

`1/(x^2-4) - 1/(x^2+4) = ((x^2+4)-(x^2-4))/(x^4-16) = 8/(x^4-16)`

So:

`1/(x^4-16) = 1/(8(x^2-4))-1/(8(x^2+4))`

Then note that:

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

So:

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

And:

`1/(8(x^2-4)) = 1/(32(x-2))-1/(32(x+2))`

Putting it all together:

`1/(x^4-16) = 1/(32(x-2))-1/(32(x+2))-1/(8(x^2+4))`