Question

How do you write the partial fraction decomposition of the rational expression `(8x^2 - 4x - 8)/(x^4 + 2x^3)`?

Answer 1

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

Explanation:

We have: `(8x^2 - 4x - 8)/(x^4 + 2x^3)=(8x^2 - 4x - 8)/(x^3(x+ 2))`

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

`color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax+B(x+2))/(x(x+2))+C/x^2+D/x^3`

`color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax^2+Bx(x+2)+C(x+2))/(x^2(x+2))+D/x^3`

`color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax^3+Bx^2(x+2)+Cx(x+2)+D(x+2))/(x^3(x+2))`

`8x^2 - 4x - 8=Ax^3+Bx^2(x+2)+Cx(x+2)+D(x+2)`

Putting in `x=0` gives us:
`-8=D(2)`
`D=-4`

`8x^2 - 4x - 8=Ax^3+Bx^2(x+2)+Cx(x+2)-4(x+2)`

Now putting in `x=-2` gives us:
`8(-2)^2 - 4(-2) - 8=A(-2)^3+B(-2)^2(-2+2)+C(-2)(-2+2)-4(-2+2)`
`32=-8A`
`A=-4`

`8x^2 - 4x - 8=-4x^3+Bx^2(x+2)+Cx(x+2)-4(x+2)`

There are no numbers that can give us just `B` or `C`, so we must find one in terms of the other, and so we shall put in `x=1` to get:
`-4=-4+3B+3C-12`
`12=3(B+C)`
`B=4-C` (we'll shall put this back in the original equation to help find `C`.

`8x^2 - 4x - 8=-4x^3+(4-C)x^2(x+2)+Cx(x+2)-4(x+2)`

We need to put in a random value, i.e. `x=2`

`8(2)^2 - 4(2)- 8=-4(2)^3+(4-C)(2)^2((2)+2)+C(2)((2)+2)-4((2)+2)`

`16=8c+16(4-C)-48`

`64=16(4-C)`

`4=4-C`

`C=0`

Remember that `B=4-C`

`B=4-0=4`

`(8x^2 - 4x - 8)/(x^3(x+ 2))=-4/(x+2)+4/x+0/x^2-4/x^3`

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