Question

How do you express `x^4/(x-1)^3` in partial fractions?

Answer 1

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

Explanation:

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

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

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

`color(white)(x^4/(x-1)^3) = x + ((3x^3-9x^2+9x-3)+(6x^2-8x+3))/(x^3-3x^2+3x-1)`

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

`color(white)(x^4/(x-1)^3) = x + 3+A/(x-1)+B/(x-1)^2+C/(x-1)^3`

`color(white)(x^4/(x-1)^3) = x + 3+(A(x-1)^2+B(x-1)+C)/(x-1)^3`

`color(white)(x^4/(x-1)^3) = x + 3+(A(x^2-2x+1)+B(x-1)+C)/(x-1)^3`

`color(white)(x^4/(x-1)^3) = x + 3+(Ax^2+(-2A+B)x+(A-B+C))/(x-1)^3`

Equating coefficients, we find:

`{ (A = 6), (-2A+B=-8), (A-B+C=3) :}`

Hence:

`{ (A = 6), (B=4), (C=1) :}`

So:

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