Chapter 11.5, Problem 22AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{2x+4}{x^3-1}$

To determine

To find: The partial fraction decomposition of the given rational function.

Answer to Problem 22AYU

Solution:

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

Explanation of Solution

Given:

$\frac{2x + 4}{x^3 - 1}$

Formula used:

a. If $Q(x) = (x - a_1)\dots \dots .(x - a_n)$ then partial fraction decomposition of $\text{P}\left(\text{x}\right)\text{/Q}\left(\text{x}\right)$ takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a_1\right)} + \frac{A_2}{\left(x - a_2\right)} + \cdots \frac{A_n}{\left(x - a_n\right)}$

where $A_1, A_2\dots A_n$ are to be determined.

b. If $Q\left(x\right)$ contains a non-repeated irreducible quadratic factor of the form $a{x}^2 + bx + c$, then, in the partial fraction decomposition of $P\left(x\right)\text{/}Q\left(x\right)$, takes the form:

$\frac{Ax + B}{a{x}^2 + bx + c}$

where the numbers $\text{A}$ and $\text{B}$ are to be determined.

Calculation:

The denominator can be factored as

$x^3 - 1 = \left(x - 1\right)\left(x^2 + x + 1\right)$

The given partial fraction decomposition takes the form

$\frac{2x + 4}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + c}{\left(x^2 + x + 1\right)}$   -----Eq (1)

$\begin{array}{c}2x + 4 = A\left(x^2 + x + 1\right) + \left(Bx + C\right)\left(x - 1\right)\\ = A{x}^2 + Ax + A - Bx + B{x}^2 + Cx - C\end{array}$

Equating the coefficients of $x^2$, $x$ and constant we get

$\begin{array}{c}A + B = 0\\ A - B + C = 2\\ A - C = 4\end{array}$

Solving for $A, B, C$

$\begin{array}{l}A = 2\\ B = - 2\\ C = - 2\end{array}$

From Eq (1), the partial fraction decomposition of

$\frac{2x + 4}{x^3 - 1}$ is $\frac{2}{\left(x - 1\right)} - \frac{2\left(x + 1\right)}{\left(x^2 + x + 1\right)}$.