Question

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

Answer 1

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

Explanation:

begin by factorising the denominator

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

`rArr cancel(x)/(cancel(x)(x-3)(x+2)) =1/((x-3)(x+2))`

`rArr 1/((x-3)(x+2)) = A/(x-3) + B/(x+2)`

multiply through the equation by (x-3)(x+2)

hence 1 = A(x+2) + B(x-3) ............(*)

[note that x=-2 and x=3 will make the terms with A and B equal to zero.]

let x = -2 in (*): 1 = -5B `rArr B = -1/5`

let x = 3 in(*) : 1 = 5A `rArr A = 1/5`

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