Question

How do you write `x^4/(x-1)^3` as a partial fraction decomposition?

Answer 1

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

Explanation:

Since the degree of the numerator is greater than the degree of the denominator, we perform a long division

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

`x^4` `color(white)(aaaaaaaaaaaaaaaaaaa)` `∣` `x^3-3x^2+3x-1`

`x^4-3x^3+3x^2-x` `color(white)(aaaaa)` `∣` `x+3`

`0+3x^3-3x^2+x`

`color(white)(aaa)` `3x^3-9x^2+9x-3`

`color(white)(aaaaa)` `0+6x^2-8x+3`

So we get

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

now we form the partial fraction decomposition

`(6x^2-8x+3)/(x-1)^3=A/(x-1)+B/(x-1)^2+C/(x-1)^3`

`(6x^2-8x+3)/(x-1)^3=(A(x-1)^2+B(x-1)+C)/(x-1)^3`

So `6x^2-8x+3=A(x-1)^2+B(x-1)+C`

Let `x=1` then `1=C`

Compare the coefficients of `x^2`

`6=A`

Let `x=0`, then `3=A-B+C`

`B=6+1-3=4`

So the final result is

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