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

##### 1 Answer

Feb 5, 2016

# 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)) #