Chapter 11.5, Problem 30AYU

To determine

To find:

The partial fraction decomposition of the given rational function.

Answer to Problem 30AYU

Solution:

$\frac{ - 18}{13x} + \frac{\left(\frac{31}{13}\right)x - \frac{89}{13}}{\left(x^2 + 3x + 13\right)}$

Explanation of Solution

Given:

$\frac{x^2 - 11x - 18}{x\left(x^2 + 3x + 13\right)}$

Formula used:

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

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

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

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

Calculation:

The given partial fraction decomposition takes the form

$\frac{x^2 - 11x - 18}{x\left(x^2 + 3x + 13\right)} = = \frac{A}{x} + \frac{Bx + c}{\left(x^2 + 3x + 13\right)}$

$= >x^2 - 11x - 18 = A\left(x^2 + 3x + 13\right) + \left(Bx + C\right)x$   -----…Eq(1)

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

$\begin{array}{cc}{x}^2, & = >A + B = 1\end{array}$   -----Eq(2)

$\begin{array}{cc}x, & = >3A + C = - 11\end{array}$   -----Eq(3)

$\text{constant} = >13A = 18$   -----Eq(4)

Solving Eq(2), Eq(3), Eq(4) for $A$, $B$, $C$, $D$ we get

$\text{A} = - 18/13, \text{B = 31/13,} \text{C = 89/13}$

From Eq (1), the partial fraction decomposition of $\frac{x^2 - 11x - 18}{x\left(x^2 + 3x + 13\right)}$ is $\frac{ - 18}{13x} + \frac{\left(\frac{31}{13}\right)x\frac{89}{13}}{\left(x^2 + 3x + 13\right)}$.