Chapter 7.4, Problem 32E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 32E

  $\frac{x+6}{x^3-3x^2-4x+12}=\boxed{\frac{9}{5\left(x-3\right)}+\frac{1}{5\left(x+2\right)}-\frac{2}{x-2}}$

Explanation of Solution

Given information: $\frac{x+6}{x^3-3x^2-4x+12}$

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:

Factor the GCF of first two terms and GCF of last two terms of the denominator of the given rational expression

  $\frac{x+6}{x^3-3x^2-4x+12}=\frac{x+6}{x^2\left(x-3\right)-4\left(x-3\right)}$

Factor the common binomial of the denominator to factor the denominator further

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

Rewrite the expression $x^2-4$

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

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

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

Write one partial fraction for each of those factors

  $\frac{x+6}{\left(x-3\right)\left(x+2\right)\left(x-2\right)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{C}{x-2}$

Multiply through by the LCD $\left(x-3\right)\left(x+2\right)\left(x-2\right)$ so we no longer have fractions

  $\begin{array}{l}\frac{\left(x+6\right)\cdot \left(x-3\right)\left(x+2\right)\left(x-2\right)}{\left(x-3\right)\left(x+2\right)\left(x-2\right)}=\frac{A\cdot \left(x-3\right)\left(x+2\right)\left(x-2\right)}{x-3}+\frac{B\cdot \left(x-3\right)\left(x+2\right)\left(x-2\right)}{x+2}\\ +\frac{C\cdot \left(x-3\right)\left(x+2\right)\left(x-2\right)}{x-2}\end{array}$

Cancel common factors throughout the equation

  $x+6=A\left(x+2\right)\left(x-2\right)+B\left(x-3\right)\left(x-2\right)+C\left(x-3\right)\left(x+2\right)$

Expand the expression in the right side of the equation

  $x+6=A\left(x^2-4\right)+B\left(x^2-5x+6\right)+C\left(x^2-x-6\right)$

Distribute in right side of the equation

  $x+6=A{x}^2-4A+B{x}^2-5Bx+6B+C{x}^2-Cx-6C$

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

  $x+6=A{x}^2+B{x}^2+C{x}^2-5Bx-Cx-4A+6B-6C$

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

  $0x^2+1x+6=\left(A+B+C\right)x^2+\left(-5B-C\right)x+\left(-4A+6B-6C\right)$

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

  $A+B+C=0$

  $-5B-C=1$

  $-4A+6B-6C=6$

Label the equation we got

  $A+B+C=0$  ▶ 1^st equation

  $-5B-C=1$  ▶ 2^nd equation

  $-4A+6B-6C=6$  ▶ 3^rd equation

In order to eliminate variable A, multiply 4 with the 1^st equation and add the resulting equation to the 3^rd equation

  $\begin{array}{l}4A+4B+4C=0\\ \underset{\_}{-4A+6B-6C=6}\\ 10B-2C=6\end{array}$

Label the resulting equation as 4^th equation

  $10B-2C=6$  ▶ 4^th equation

In order to eliminate variable C, multiply -2 to the 2^nd equation and add the result to the 4^th equation

  $\begin{array}{l}10B+2C=-2\\ \underset{\_}{10B-2C=6}\\ 20B=4\end{array}$

Solve the equation $20B=4$

• By dividing 20 on both sides, then
• Simplifying fraction on both sides of the equation

  $\frac{20B}{20}=\frac{4}{20}$

  $B=\frac{1}{5}$

Substitute 1/5 for B in the 2^nd equation and Solve the equation for C

• Simplify fraction in left side of the equation
• By adding 1 on both sides, and then
• Dividing by -1 on both sides

  $-5\left(\frac{1}{5}\right)-C=1$

  $-1-C=1$

  $-1-C+1=1+1$

  $-C=2$

  $\frac{-C}{-1}=\frac{2}{-1}$

  $C=-2$

Substitute -2 for C and 1/5 for B in the 1^st equation and solve the equation for A

• By adding fractions in the left side of the equation, and then
• Adding 9/5 on both sides of the equation

  $A+\frac{1}{5}-2=0$

  $A+\frac{1}{5}-\frac{2\cdot 5}{1\cdot 5}=0$

  $A+\frac{1}{5}-\frac{10}{5}=0$

  $A-\frac{9}{5}=0$

  $A-\frac{9}{5}+\frac{9}{5}=0+\frac{9}{5}$

  $A=\frac{9}{5}$

Substitute the value of A, B, and C in the equation $\frac{x+6}{\left(x-3\right)\left(x+2\right)\left(x-2\right)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{C}{x-2}$

  $\frac{x+6}{\left(x-3\right)\left(x+2\right)\left(x-2\right)}=\frac{9/5}{x-3}+\frac{1/5}{x+2}+\frac{-2}{x-2}$

Conclusion:

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