Q. 32 - Precalculus Enhanced with Graphing Utilities

Question

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

$\frac{1}{(2x+3)(4x-1)}$

Step-by-Step Solution

Answer

The partial fraction decomposition of a rational expression is 

$\frac{1}{(2x+3)(4x-1)}=\frac{\frac{-1}{7}}{(2x+3)}+\frac{\frac{2}{7}}{(4x-1)}$

Step 1. Given information

Given rational expression is

$\frac{1}{(2x+3)(4x-1)}$

Step 2. Partial fraction decomposition

partial fraction decomposition of a rational expression  

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n} \\ \frac{1}{(2x+3)(4x-1)}=\frac{A}{(2x+3)}+\frac{B}{(4x-1)} \cdots \left(i\right) \\ \frac{1}{(2x+3)(4x-1)}=\frac{A(4x-1)}{(2x+3)(4x-1)}+\frac{B(2x+3)}{(2x+3)(4x-1)} \\ 1=A(4x-1)+B(2x+3) \\ 0x+1=(4A+2B)x+(3B-A) \cdots \left(ii\right) \end{gathered}$$

Step 3. Values of coefficients and constants of the numerator

Compare the constants in equation ii 

$$\begin{gathered} 1=3B-A \\ A=3B-1 \end{gathered}$$

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

$$\begin{gathered} 0=4A+2B \\ 0=4(3B-1)+2B \\ B=\frac{2}{7} \end{gathered}$$

So$A=3\left(\frac{2}{7}\right)-1=\frac{-1}{7}$

Step 4. partial fraction decomposition of a rational expression

Substitute the value of A, B, and C in the equation i 

$$\begin{gathered} \frac{1}{(2x+3)(4x-1)}=\frac{A}{(2x+3)}+\frac{B}{(4x-1)} \\ \frac{1}{(2x+3)(4x-1)}=\frac{\frac{-1}{7}}{(2x+3)}+\frac{\frac{2}{7}}{(4x-1)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{1}{(2x+3)(4x-1)}=\frac{\frac{-1}{7}}{(2x+3)}+\frac{\frac{2}{7}}{(4x-1)}$