Question

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

Answer 1

`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)`