Question

How do I find the partial-fraction decomposition of `(2x^3+7x^2-2x+6)/(x^4+4)`?

Answer 1

First factor the denominator to find `x^4+4 = (x^2+2x+2)(x^2-2x+2)`

Then solve `2x^3+7x^2-2x+6 = (x^2-2x+2)(Ax+B) + (x^2+2x+2)(Cx+D)`

finding `A=0`, `B=3`, `C=2` and `D=0`.

Explanation:

`x^4+4` has no linear factors with Real coefficients since `x^4+4 >= 4 > 0` for all `x in RR`.

It will have quadratic factors with Real coefficients.

`x^4+4 = (ax^2+bx+c)(dx^2+ex+f)`

Without loss of generalisation, `a = d = 1`, so

`x^4+4 = (x^2+bx+c)(x^2+ex+f)`

Then looking at the coefficient of `x^3`, we have `e = -b`, so

`x^4+4 = (x^2+bx+c)(x^2-bx+f)`

Then looking at the coefficient of `x` we have `b(f-c) = 0`, so either `b=0` or `c = f`.

If `b=0` then `c+f = 0` and `cf = -c^2 = 4`, which has no Real solutions (it has solutions `c = 2i`, `f = -2i`).

If `c = f` and `cf = 4` then `c=f=+-2`. Then the coefficient of `x^2` tells us `f+c-b^2 = 0`. Now if `b` is Real, then `b^2 >= 0`, so `f+c >= 0`, hence `f=c=2`. Then `b^2 = 4`, so `b = +-2`.

So we have `x^4+4 = (x^2+2x+2)(x^2-2x+2)`

Now attempt to solve:

`(2x^3+7x^2-2x+6)/(x^4+4) = (Ax+B)/(x^2+2x+2) + (Cx+D)/(x^2-2x+2)`

If we multiply through by `(x^4+4) = (x^2+2x+2)(x^2-2x+2)` then this becomes:

`2x^3+7x^2-2x+6`

`= (x^2-2x+2)(Ax+B) + (x^2+2x+2)(Cx+D)`

`= (A+C)x^3+(B-2A+D+2C)x^2+2(A-B+C+D)x+2(B+D)`

Equating coefficients we get:

(i) `A+C = 2`
(ii) `B-2A+D+2C = 7`
(iii) `2(A-B+C+D) = -2`
(iv) `2(B+D) = 6`

From (i) and (iv) we get:

(v) `C = 2 - A` and
(vi) `D=3-B`

Substitute these into (ii) to get:

`color(red)(cancel(color(black)(B)))-2A+3-color(red)(cancel(color(black)(B)))+4-2A=7`

Hence `A=0`. Then from (v) `C = 2`.

Substitute `A=0`, `C=2` and `D=3-B` into (iii) to get:

`2(0-B+2+3-B) = -2`

Divide both sides by `2` to find:

`5-2B = -1`

Hence `B=3`. Then from (vi) `D=0`.

So

`(2x^3+7x^2-2x+6)/(x^4+4) = 3/(x^2+2x+2) + (2x)/(x^2-2x+2)`

Answer 2

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

`(2x^3+7x^2-2x+6)/(x^4+4) = 3/(x^2+2x+2) + (2x)/(x^2-2x+2)`

Explanation:

Here is a quick way to find the quadratic factors of `x^4+4` using complex arithmetic ...

`sqrt(-4) = 2i`

`sqrt(2i) = i + 1`

If `x_1` is a root of `x^4+4 = 0`, then so are `-x_1`, `bar(x_1)` and `bar(-x_1)`, where `bar(x_1)` means the complex conjugate. This four-fold symmetry occurs due to the only power of `x` being `x^4`.

Here are the roots in the complex plane:

graph{((x-1)^2+(y-1)^2 - 0.01)((x+1)^2+(y-1)^2 - 0.01)((x-1)^2+(y+1)^2 - 0.01)((x+1)^2+(y+1)^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

So `x^4+4 = (x-1-i)(x-1+i)(x+1-i)(x+1+i)`

Now pick pairs of factors to multiply to get Real coefficients.

To do this, pick the ones which are complex conjugates:

`(x+1-i)(x+1+i) = ((x+1)-i)((x+1)+i)`

`= (x+1)^2-i^2 = x^2+2x+1+1 = x^2+2x+2`

and

`(x-1-i)(x-1+i) = ((x-1)-i)((x-1)+i)`

`= (x-1)^2-i^2 = x^2-2x+1+1 = x^2-2x+2`

For the rest of the partial fraction decomposition, see the other answers.