Question

How do you express `x^3/(x^2 + 4x + 3 )` in partial fractions?

Answer 1

`27/(2(x+3)) -1/(2(x+1))`

Explanation:

the first step is to factor the denominator

`x^2 + 4x + 3 = (x+1)(x+3)`

since these factors are linear , the denominators of the partial fractions will be constants say A and B.

`x^3/((x+1)(x+3)) = A/(x+1) + B/(x+3)`

multiply through by (x+1)(x+3)

hence : `x^3 = A(x+3) + B(x+1) ..............................(1)`

The aim now is to find the values of A and B . Note that if x = -3 the term with A will be zero and if x = -1 the term with B will be zero. This is the starting point for finding the values.

let x = - 3 in (1) : -27 = -2B `rArr B = 27/2`

let x = -1 in (1) : -1 = 2A `rArr A = -1/2`

`rArr x^3/(x^2+4x+3) = 27/(2(x+3)) - 1/(2(x+1))`