Chapter 11.5, Problem 37AYU

To determine

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

Answer to Problem 37AYU

Solution:

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

Explanation of Solution

Given:

$\frac{7x + 3}{x^3 - 2x^2 - 3x}$

Formula used:

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

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

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

Calculation:

The denominator of the rational function $x^3 - 2x^2 - 3x$ can be factored as $x(x + 1)(x - 3)$

The given partial fraction decomposition takes the form

$\begin{array}{l}\frac{7x + 3}{x^3 - 2x^2 - 3x} = \frac{A}{x} + \frac{B}{x + 1} + \frac{c}{x - 3}\\ \Rightarrow 7x + 3 = A(x + 1)(x - 3) + Bx(x - 3) + Cx(x + 1)\end{array}$   -----Eq(1)

Equating the co-efficients on both sides in Eq(1),

$x^2, \Rightarrow A + B + C = 0$   -----Eq(2)

$x, \Rightarrow - 2A - 3B + C = 7$   -----Eq(3)

$constant \Rightarrow - 3A = 3$   -----Eq(4)

Solving for $A,B, C$ we get

$A = - 1, B = - 1 C = 2$

The partial fraction decomposition of $\frac{7x + 3}{x^3 - 2x^2 - 3x}$ is $- \frac{1}{x} - \frac{1}{x + 1} + \frac{2}{x - 3}$.