Chapter 11.5, Problem 18AYU

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

To determine

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

Answer to Problem 18AYU

Solution:

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

Explanation of Solution

Given:

$\frac{3x}{\left(x + 2\right)\left(x - 4\right)}$

Formula used:

If $Q(x) = (x - a_1)\mathrm{...}\mathrm{....}(x - a_n)$ then partial fraction decomposition of $\text{P(x)}/\text{Q(x)}$ 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 $A_1,A_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

The given partial fraction can be written as

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

$\begin{array}{lll}3x\hfill & = \hfill & A\left(x - 4\right) + B\left(x + 2\right)\hfill \\ \hfill & = \hfill & Ax - 4A + Bx + 2B\hfill \end{array}$

Equating the coefficients of $x$ and constant we get

$\begin{array}{lll}\hfill A + B& = \hfill & 3\hfill \\ - 4A + 2B\hfill & = \hfill & 0\hfill \end{array}$

Solving for $A, B$

$\begin{array}{lll}A\hfill & = \hfill & 1\hfill \\ B\hfill & = \hfill & 2\hfill \end{array}$

From Eq (1), the partial fraction decomposition of $\frac{3x}{\left(x + 2\right)\left(x - 4\right)}$ is $\frac{1}{x + 2} + \frac{2}{x - 4}$.