Chapter 7.4, Problem 31E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 31E

  $\frac{x}{x^3-x^2-2x+2}=\boxed{\frac{x+2}{x^2-2}-\frac{1}{x-1}}$

Explanation of Solution

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

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

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

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

  $\frac{x}{x^2\left(x-1\right)-2\left(x-1\right)}=\frac{x}{\left(x-1\right)\left(x^2-2\right)}$

Write one partial fraction for each of those factors

  $\frac{x}{\left(x-1\right)\left(x^2-2\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2-2}$

Multiply through by the LCD $\left(x-1\right)\left(x^2-2\right)$ so we no longer have fractions

  $\frac{x\cdot \left(x-1\right)\left(x^2-2\right)}{\left(x-1\right)\left(x^2-2\right)}=\frac{A\cdot \left(x-1\right)\left(x^2-2\right)}{x-1}+\frac{\left(Bx+C\right)\cdot \left(x-1\right)\left(x^2-2\right)}{x^2-2}$

Cancel common factors throughout the equation

  $x=A\left(x^2-2\right)+\left(Bx+C\right)\left(x-1\right)$

Expand the expression in the right side of the equation

  $x=A{x}^2-2A+B{x}^2-Bx+Cx-C$

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

  $x=A{x}^2+B{x}^2+Cx-Bx-2A-C$

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

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

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

  $A+B=0$

  $-B+C=1$

  $-2A-C=0$

Label the equation we got

  $A+B=0$  ▶ 1^st equation

  $-B+C=1$  ▶ 2^nd equation

  $-2A-C=0$  ▶ 3^rd equation

In order to eliminate variable B, add the 1^st equation and 2^ndequation

  $\begin{array}{l}A+B=0\\ \underset{\_}{-B+C=1}\\ A+C=1\end{array}$

Label the resulting equation as 4^th equation

  $A+C=1$  ▶ 4^thequation

In order to eliminate variable C, add the 3^rdequation and 4^thequation

  $\begin{array}{l}-2A-C=0\\ \underset{\_}{A+C=1}\\ -A=1\end{array}$

Solve the equation $-A=1$

• By dividing -1 on both sides, then
• Simplifying fraction on both sides of the equation

  $\frac{-A}{-1}=\frac{1}{-1}$

  $A=-1$

Substitute -1 for A in the 1^stequation and then add 1 on both sides to get the value of B

  $-1+B=0$

  $-1+B+1=0+1$

  $B=1$

Substitute 1 for B in the 2^nd equation and then add 1 on both sides to get the value of C

  $-1+C=1$

  $-1+C+1=1+1$

  $C=2$

Substitute the value of A, B, and C in the equation $\frac{x}{\left(x-1\right)\left(x^2-2\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2-2}$

  $\frac{x}{\left(x-1\right)\left(x^2-2\right)}=\frac{-1}{x-1}+\frac{1x+2}{x^2-2}$

Conclusion:

The partial fraction decomposition of the given rational expressionis: $\boxed{\frac{x+2}{x^2-2}-\frac{1}{x-1}}$