Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 13–46, write the partial fraction decomposition of each rational expression.

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

Step-by-Step Solution

Answer

The partial fraction decomposition of a rational expression is 

$\frac{x}{(3x-2)(2x+1)}=\frac{\frac{2}{7}}{(3x-2)}+\frac{\frac{1}{7}}{(2x+1)}$

Step 1. Given information

Given rational expression is

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

Step 2. Partial fraction decomposition

partial fraction decomposition of a rational expression  

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n} \\ \frac{x}{(3x-2)(2x+1)}=\frac{A}{(3x-2)}+\frac{B}{(2x+1)} \cdots \left(i\right) \\ \frac{x}{(3x-2)(2x+1)}=\frac{A(2x+1)}{(3x-2)(2x+1)}+\frac{B(3x-2)}{(3x-2)(2x+1)} \\ x=A(2x+1)+B(3x-2) \\ x+0=(2A+3B)x+(A-2B) \cdots \left(ii\right) \end{gathered}$$

Step 3. Values of coefficients and constants of the numerator

Compare the constants in equation ii 

$$\begin{gathered} 0=A-2B \\ A=2B \end{gathered}$$

Compare the coefficient of x in equation ii  and Substitute the expression for A 

$$\begin{gathered} 1=2A+3B \\ 1=2\left(2B\right)+3\left(B\right) \\ B=\frac{1}{7} \end{gathered}$$

so

$A=2\left(\frac{1}{7}\right)=\frac{2}{7}$

Step 4. partial fraction decomposition of a rational expression

Substitute the value of A and B in the equation i 

$$\begin{gathered} \frac{x}{(3x-2)(2x+1)}=\frac{A}{(3x-2)}+\frac{B}{(2x+1)} \\ \frac{x}{(3x-2)(2x+1)}=\frac{\frac{2}{7}}{(3x-2)}+\frac{\frac{1}{7}}{(2x+1)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{x}{(3x-2)(2x+1)}=\frac{\frac{2}{7}}{(3x-2)}+\frac{\frac{1}{7}}{(2x+1)}$