Chapter 11.5, Problem 44AYU

To determine

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

Answer to Problem 44AYU

Solution:

$\frac{8}{5\left(x + 2\right)} + \frac{4}{5\left(2x - 1\right)}$

Explanation of Solution

Given:

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

Formula used:

If $Q(x) = (x - a_1)\mathrm{...} \mathrm{....}(x - a_n)$ 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 $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

The denominator $2x^2 + 3x - 2$ can be factored as $\left(x + 2\right)\left(2x - 1\right)$

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

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

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

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

$\begin{array}{ll}x,\hfill & \Rightarrow 2A + B = 4\hfill \end{array}$   -----Eq(2)

$constant\Rightarrow - A + 2B = 0$   -----Eq(3)

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

$\text{A} = 8/5, \text{B} = 4/5$

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