# How do you write the partial fraction decomposition of the rational expression # (x+4)/((x+1)(x-2)^2 )#?

##### 1 Answer

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

#### Explanation:

firstly note that

# x-2)^2#

has factors (x-2) and# (x-2)^2 # the factors of the denominator are linear hence the numerators are constants.

# rArr (x+4)/((x+1)(x-2)^2 ) = A/(x+1) + B/(x-2) + C/(x-2)^2 # now multiply through by

#(x+1)(x-2)^2 # hence : x+4 =

#A(x-2)^2 + B(x+1)(x-2) + C(x+1)....................(1)# the aim now is to find values for A , B and C . Note that if x = 2 then the terms with A and B will be zero , and if x = -1 the terms with B and C will be zero. This is the starting point in finding values.

let x = 2 in (1) :

# 6 = 3C rArr C = 2# let x = -1 in (1) : -3 =

# 9A rArr A = 1/3 #

any value for x may be chosen to find B.let x = 0 in (1) : 4 = 4A - 2B + C

# rArr 2B = 4A + C - 4 # hence 2B

# = 4/3 + 2 - 4 = -2/3 rArr B = -1/3 #

#rArr(x+4)/((x+1)(x-2)^2) = 1/3(x+1) -1/3(x-2) + 2/(x-2)^2 #