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

Feb 18, 2016

$\frac{27}{2 \left(x + 3\right)} - \frac{1}{2 \left(x + 1\right)}$

#### Explanation:

the first step is to factor the denominator

${x}^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right)$

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

${x}^{3} / \left(\left(x + 1\right) \left(x + 3\right)\right) = \frac{A}{x + 1} + \frac{B}{x + 3}$

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

hence : ${x}^{3} = A \left(x + 3\right) + B \left(x + 1\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

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 $\Rightarrow B = \frac{27}{2}$

let x = -1 in (1) : -1 = 2A $\Rightarrow A = - \frac{1}{2}$

$\Rightarrow {x}^{3} / \left({x}^{2} + 4 x + 3\right) = \frac{27}{2 \left(x + 3\right)} - \frac{1}{2 \left(x + 1\right)}$