Question

How do you express `(4x^3-22x^2+33x-6)/((x-3)^3(x+2))` in partial fractions?

Answer 1

The decomposition is `=(3/5)/(x-3)^3+(42/25)/(x-3)^2+(308/125)/(x-3)+(192/125)/(x+2)`

Explanation:

The decomposition is `(4x^3-22x^2+33x-6)/((x-3)^3(x+2))=A/(x-3)^3+B/(x-3)^2+C/(x-3)+D/(x+2)`
`=A(x+2)+B(x-3)(x+2)+C(x-3)^2(x+2)+D(x-3)^3`

`4x^3-22x^2+33x-6=A(x+2)+B(x-3)(x+2)+C(x-3)^2(x+2)+D(x-3)^3`
let `x=3` `=>` `5A=3` `=>` `A=3/5`
let `x=-2` `=>` `-125D=-192` `=>` `D=192/125`
coefficients of `x^3` `=>` `C+D=4` `=>` `C=4-192/125=308/125`
and putting `x=0`, we get `-6=2A-6B+18C-27D`

or `=>6B=6+6/5+18*308/25-27*192/125`

= `36/5+9(616/125-576/25)`

= `36/5+360/125`

= `180/25+72/25=252/25`

`=>B=42/25`