Question

How do you write the partial fraction decomposition of the rational expression `(5x-1)/(x^2-x-2)`?

Answer 1

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

Explanation:

General form to break into partial fractions is
`(ax+b)/(cx^2+dx+e)=A/(x+C)+B/(x+D)`

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

`(5x-1)/((x-2)(x+1))=A/(x-2)+B/(x+1)=(A(x+1)+B(x-2))/((x-2)(x+1))`

`(5x-1)/((x-2)(x+1))=(Ax+A+Bx-2B)/((x-2)(x+1))`

we can equate the coefficients of numerators of both sides
`A+B=5,A-2B=-1`
Subtracting both gives `3B=6`
or B = 2
And A=3
thus

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