Chapter 11.5, Problem 43AYU

To determine

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

Answer to Problem 43AYU

Solution:

$\frac{4}{7\left(x - 3\right)} - \frac{8}{7\left(2x + 1\right)}$

Explanation of Solution

Given:

$\frac{4}{2x^2 - 5x - 3}$

Formula used:

If $Q\left(x\right) = \left(x - a_1\right)\dots \dots .\left(x - a_n\right)$ 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 $2x^2 - 5x - 3$ can be factored as $\left(x - 3\right)\left(2x + 1\right)$

The denominator has repeated linear irreducible quadratic factor. Therefore the partial fraction decomposition takes the form

$\frac{4}{2x^2 - 5x - 3} = \frac{4}{\left(x - 3\right)\left(2x + 1\right)} = \frac{A}{\left(x - 3\right)} + \frac{B}{\left(2x + 1\right)}$

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

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

$\begin{array}{ll}x,\hfill & \Rightarrow 2A + B = 0\hfill \end{array}$   -----Eq(2)

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

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

$\text{A} = 4/7, \text{B} = - 8/7$

The partial fraction decomposition of $\frac{4}{2x^2 - 5x - 3}$ is $\frac{4}{7\left(x - 3\right)} - \frac{8}{7\left(2x + 1\right)}$