Chapter 11.5, Problem 34AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{x^2-x-8}{\left(x+1\right)\left(x^2+5x+6\right)}$

To determine

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

Answer to Problem 34AYU

Solution:

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

Explanation of Solution

Given:

$\frac{x^2 - x - 8}{\left(x + 1\right)\left(x^2 + 5x + 6\right)}$

Formula used:

If $Q\left(x\right) = \left(x - a_1\right)\dots \dots .\left(x \text{-} a_n\right)$ 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.

Calculation:

The denominator $\left(x + 1\right)\left(x^2 + 5x + 6\right)$ can be rewritten as $\left(x + 1\right)\left(x + 2\right)\left(x + 3\right)$.

The given partial fraction decomposition takes the form

$\frac{x^2 - x - 8}{\left(x + 1\right)\left(x^2 + 5x + 6\right)} = \frac{x}{\left(x + 1\right)\left(x + 2\right)\left(x + 3\right)} = \frac{A}{\left(x + 1\right)} + \frac{B}{\left(x + 2\right)} + \frac{C}{\left(x + 3\right)}$

$= >x^2 - x - 8 = A\left(x + 2\right)\left(x + 3\right) + B\left(x + 1\right)\left(x + 3\right) + C\left(x + 1\right)\left(x + 2\right)$   ----- Eq (1)

Substituting in Eq(1), $\text{x} = \text{-} 1 = >\text{A} = \text{-} 3$

Substituting in Eq(1), $\text{x} = \text{-} 2 = >\text{B} = 2$

Substituting in Eq(1), $\text{x} = \text{-} 3 = >\text{C} = 2$

Hence, the partial fraction decomposition of $\frac{x^2 - x - 8}{\left(x + 1\right)\left(x^2 + 5x + 6\right)}$ is $- \frac{3}{\left(x + 1\right)} + \frac{2}{\left(x + 2\right)} + \frac{2}{\left(x + 3\right)}$.