Chapter 7.4, Problem 26E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 26E

  $\frac{x+2}{x\left(x^2-9\right)}=\boxed{\frac{5}{18\left(x-3\right)}-\frac{1}{18\left(x+3\right)}-\frac{2}{9x}}$

Explanation of Solution

Given information: $\frac{x+2}{x\left(x^2-9\right)}$

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:

Rewrite the denominator

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

Use the algebraic identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$ to factor further

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

Write one partial fraction for each of those factors

  $\frac{x+2}{x\left(x^2-9\right)}=\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x-3}$

Multiply through by the LCD $x\left(x+3\right)\left(x-3\right)$ so we no longer have fractions

  $\frac{\left(x+2\right)\cdot x\left(x+3\right)\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{A\cdot x\left(x+3\right)\left(x-3\right)}{x}+\frac{B\cdot x\left(x+3\right)\left(x-3\right)}{x+3}+\frac{C\cdot x\left(x+3\right)\left(x-3\right)}{x-3}$

Cancel common factors throughout the equation

  $x+2=A\left(x+3\right)\left(x-3\right)+Bx\left(x-3\right)+Cx\left(x+3\right)$

Expand the expression in the right side of the equation

  $x+2=A\left(x^2-3^2\right)+B\left(x^2-3x\right)+C\left(x^2+3x\right)$

Distribute A, B, & C in right side of the equation

  $x+2=A{x}^2-9A+B{x}^2-3Bx+C{x}^2+3Cx$

Group $x^2,x,\&$ constant terms in right side of the equation

  $x+2=A{x}^2+B{x}^2+C{x}^2-3Bx+3Cx-9A$

Factor $x^2,x,\&$ constant terms in right side of the equation

  $0x^2+1x+2=\left(A+B+C\right)x^2+\left(-3B+3C\right)x+-9A$

Equate the coefficient of $x^2,x,\&$ constant in both sides of the equation

  $A+B+C=0$

  $-3B+3C=1$

  $-9A=2$

Label the equation we got

  $A+B+C=0$  ▶ 1^st equation

  $-3B+3C=1$  ▶ 2^nd equation

  $-9A=2$  ▶ 3^rd equation

Solve the 3^rd equation $-9A=2$

• By dividing -9 on both sides, and
• Simplifying the fraction on both sides

  $-9A/-9=2/-9$

  $A=-2/9$

Substitute -2/9 for A in 1^st equation and then add 2/9 on both sides to get B and C alone in one side to get 4^th equation.

  $-\frac{2}{9}+B+C=0$

  $-\frac{2}{9}+B+C+\frac{2}{9}=0+\frac{2}{9}$

  $B+C=2/9$  ▶ 4^th equation

Multiply -3 to the 4^th equation and add the result 2^nd equation

  $\begin{array}{l}-3B-3C=-\frac{2}{3}\\ \underset{\_}{-3B+3C=1}\\ -6B=-\frac{2}{3}+1\end{array}$

Solve the $-6B=-\frac{2}{3}+1$

• By adding the fractions in the right side of the equation
• By dividing -6 on both sides,
• Simplifying the fraction on both sides, and
• Flip the second fraction and multiply with the first fraction in right side of the equation

  $-6B=-\frac{2}{3}+\frac{3}{3}$

  $-6B=\frac{1}{3}$

  $-6B÷-6=\frac{1}{3}÷-6$

  $B=\frac{1}{3}\times -\frac{1}{6}$

  $B=-\frac{1}{18}$

Substitute -1/18for B in the 4^th equation and then add1/18 on both sides to get the value of C

  $\frac{-1}{18}+C=\frac{2}{9}$

  $\frac{-1}{18}+C+\frac{1}{18}=\frac{2}{9}+\frac{1}{18}$

  $C=\frac{2\cdot 2}{9\cdot 2}+\frac{1}{18}$

  $C=\frac{4}{18}+\frac{1}{18}$

  $C=\frac{5}{18}$

Substitute the value of A, B, and C in the equation $\frac{x+2}{x\left(x^2-9\right)}=\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x-3}$

  $\frac{x+2}{x\left(x^2-9\right)}=\frac{-2/9}{x}+\frac{-1/18}{x+3}+\frac{5/18}{x-3}$

Conclusion:

The partial fraction decomposition of the given rational expressionis: $\boxed{\frac{5}{18\left(x-3\right)}-\frac{1}{18\left(x+3\right)}-\frac{2}{9x}}$