Question

How do you write the partial fraction decomposition of the rational expression `(X^2-5x+6)/(X^3-X^2+2X)`?

Answer 1

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

Explanation:

Factor out an `x` in the denominator

`(x^2-5x+6)/(x(x^2-x+2))=A/x+(Bx+C)/(x^2-x+2)`

Multiply both sides by `x(x^2-x+2)`

`x^2-5x+6=A(x^2-x+2)+x(Bx+C)`

Distribute on right hand side

`x^2-5x+6=Ax^2-Ax+2A+Bx^2+Cx`

Factor right hand side

`x^2-5x+6=x^2(A+B)+x(-A+C)+2A`

Equate coefficients on the left hand side with the right

`2A=6`

`-A+C=-5`

`A+B=1`

Solving we get

`A=3`

`B=-2`

`C=-2`

Putting it all together

`(x^2-5x+6)/(x^3-x^2+2x)=3/x+(-2x-2)/(x^2-x+2)`

Pull out negative in `-2x-2`

`(x^2-5x+6)/(x^3-x^2+2x)=3/x-(2x+2)/(x^2-x+2)`

Factor out a `2` in `2x+2`

`(x^2-5x+6)/(x^3-x^2+2x)=3/x-(2(x+1))/(x^2-x+2)`