Chapter 7.4, Problem 22E

To determine

To write : the partial fraction decomposition ofthe rational expression

Answer to Problem 22E

  $\frac{1}{4x^2-9}=\boxed{\frac{1}{6\left(2x-3\right)}-\frac{1}{6\left(2x+3\right)}}$

Explanation of Solution

Given information: $\frac{1}{4x^2-9}$

Concept Involved:

Decomposition of $N\left(x\right)/D\left(x\right)$ into Partial Fractions:

1. Divide when improper: When $N\left(x\right)/D\left(x\right)$ is an improper fraction [degree ofN(x) = degree of D(x)], divide the denominator into the numerator to obtain

$\frac{N\left(x\right)}{D\left(x\right)}=\left(polynomial\right)+\frac{N_1\left(x\right)}{D\left(x\right)}$

and apply Steps 2, 3, and 4 to the proper rational expression $\frac{N_1\left(x\right)}{D\left(x\right)}$

Note that $N_1\left(x\right)$ is the remainder from the division of $N\left(x\right)$ by $D\left(x\right)$ .

2. Factor the denominator: Completely factor the denominator into factors of the form ${\left(px+q\right)}^m$ and ${\left(a{x}^2+bx+c\right)}^n$ where $\left(a{x}^2+bx+c\right)$ is irreducible
3. Linear factors: For each factor of the form ${\left(px+q\right)}^m$ , the partial fraction decomposition must include the following sum of $m$ fractions.

$\frac{A_1}{\left(px+q\right)}+\frac{A_2}{{\left(px+q\right)}^2}+\mathrm{.....}+\frac{A_m}{{\left(px+q\right)}^m}$

4. Quadratic factors: For each factor of the form ${\left(a{x}^2+bx+c\right)}^n$, the partial fraction decomposition must include the following sum of n fractions.

$\frac{B_{1}x+C_1}{a{x}^2+bx+c}+\frac{B_{2}x+C_2}{{\left(a{x}^2+bx+c\right)}^2}+\mathrm{.....}+\frac{B_{n}x+C_n}{{\left(a{x}^2+bx+c\right)}^n}$

Calculation:

Description	Steps
Rewrite the denominator of the given rational expression	$\begin{array}{l}\frac{1}{4x^2-9}=\frac{1}{2^2{x}^2-3^2}\\ =\frac{1}{{\left(2x\right)}^2-3^2}\end{array}$
Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor further	$\frac{1}{{\left(2x\right)}^2-3^2}=\frac{1}{\left(2x+3\right)\left(2x-3\right)}$
Write one partial fraction for each of those factors	$\frac{1}{\left(2x+3\right)\left(2x-3\right)}=\frac{A}{2x+3}+\frac{B}{2x-3}$
Multiply through by the LCD $\left(2x+3\right)\left(2x-3\right)$ so we no longer have fractions $\frac{1\cdot \left(2x+3\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{A\cdot \left(2x+3\right)\left(2x-3\right)}{2x+3}+\frac{B\cdot \left(2x+3\right)\left(2x-3\right)}{2x-3}$
Cancel common factors throughout the equation	$1=A\left(2x-3\right)+B\left(2x+3\right)$
Distribute A and B in right side of the equation	$1=2Ax-3A+2Bx+3B$
Group the like terms in right side of the equation	$\begin{array}{l}1=2Ax+2Bx-3A+3B\\ 0x+1=\left(2A+2B\right)x+\left(-3A+3B\right)\end{array}$
Equate the coefficient of x and divide it throughout the equation by common factor	$2A+2B=0$$\begin{array}{l}\frac{2A}{2}+\frac{2B}{2}=\frac{0}{2}\\ A+B=0\end{array}$
Label the equation as 1st equation	$A+B=0$	▶ 1st equation
Equate the constant and divide it throughout the equation by common factor	$-3A+3B=1$$\begin{array}{l}\frac{-3A}{3}+\frac{3B}{3}=\frac{1}{3}\\ -A+B=\frac{1}{3}\end{array}$
Label the equation as 2nd equation	$-A+B=1/3$	▶ 2nd equation
Add the 1st equation with the 2nd equation	$\begin{array}{l}A+B=0\\ \underset{\_}{-A+B=1/3}\\ 2B=1/3\end{array}$
In order to solve for B in the equation $2B=1/3$ Divide both sides of the equation by 2 Flip the second fraction and multiply with the first in right side Multiply numerator with numerator and denominator with denominator	$\begin{array}{l}2B÷2=1/3÷2\\ B=\frac{1}{3}\times \frac{1}{2}\\ B=\frac{1}{6}\end{array}$

Calculation (Continued):

Description	Steps
Substitute 1/6 for B in 1st equation and solve for A By subtracting 1/6 on both sides	$\begin{array}{l}A+\frac{1}{6}=0\\ A+\frac{1}{6}-\frac{1}{6}=0-\frac{1}{6}\\ A=-\frac{1}{6}\end{array}$
Substituting the values of A and B in the equation $\frac{1}{\left(2x+3\right)\left(2x-3\right)}=\frac{A}{2x+3}+\frac{B}{2x-3}$	$\frac{1}{\left(2x+3\right)\left(2x-3\right)}=-\frac{1}{6\left(2x+3\right)}+\frac{1}{6\left(2x-3\right)}$

Conclusion:

The partial fraction decomposition of the rational expression $\frac{1}{4x^2-9}$ is:

  $\boxed{\frac{1}{6\left(2x-3\right)}-\frac{1}{6\left(2x+3\right)}}$