# How do you write the partial fraction decomposition of the rational expression #(4x^2+ 21x-22)/((2x-1)(x^2 +x-42))#?

##### 1 Answer

#14/55(2x-1) + 9/65(x+7) + 248/143(x-6)#

#### Explanation:

first step is to factor the denominator.

#x^2 + x - 42 = (x+7)(x - 6) # The factors on the denominator are all linear , hence the numerators will be constants.

# (4x^2 +21x- 22)/((2x - 1)(x+7)(x-6)) = A/(2x-1) + B/(x+7) + C/(x-6)# multiply through by (2x-1)(x+7)(x-6)

#4x^2+21x- 22 = A(x+7)(x-6) + B(2x-1)(x-6) + C(2x-1)(x+7)..........(1)# the aim now is to find the values of A , B and C. Note that if x=

#1/2# then the terms with B and C will be zero. If x = -7 ,the terms with A and C will be zero and if x=6 , the terms with A and B will be zero. This is the starting point in finding A , B and C.

let x

# = 1/2 color(black)(" in (1)") #

# -21/2 =- 165/4 rArr A = 14/55# let x = -7 in (1) : 27 =

#195B rArr B = 9/65# let x = 6 in (1) : 248 = 143 C

# rArr C = 248/143 #

# rArr( 4x^2+21x-22)/((2x-1)(x^2+x-42)) = 14/55(2x-1) + 9/65(x+7) + 248/143((x-6)#