Chapter 7.4, Problem 30E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 30E

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

Explanation of Solution

Given information: $\frac{2x}{x^3-1}$

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 of the rational expression

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

Use the algebraic identify for difference of cubes $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ to rewrite the rational expression further

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

Write one partial fraction for each of those factors

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

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

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

Cancel common factors throughout the equation

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

Expand the right side of the equation

  $2x=A{x}^2+Ax+A+B{x}^2-Bx+Cx-C$

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

  $2x=A{x}^2+B{x}^2+Ax-Bx+Cx+A-C$

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

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

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

  $A+B=0$

  $A-B+C=2$

  $A-C=0$

Label the equation we got

  $A+B=0$  ▶ 1^st equation

  $A-B+C=2$  ▶ 2^nd equation

  $A-C=0$  ▶ 3^rd equation

In order to eliminate variable B, Add 1^st equation and 2^nd equation

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

Label the resulting equation as 4^th equation

  $2A+C=2$  ▶ 4^th equation

In order to eliminate the variable C, Add 3^rd equation and 4^th equation

  $\begin{array}{l}2A+C=2\\ \underset{\_}{A-C=0}\\ 3A=2\end{array}$

Solve the equation $3A=2$ for A

• By dividing 3 on both sides, and
• Simplifying fraction on both sides

$\frac{3A}{3}=\frac{2}{3}$

  $A=\frac{2}{3}$

Substitute 2/3 for A in 1^st equation and solve for B

• By subtracting 2/3 on both sides

  $\frac{2}{3}+B=0$

  $\frac{2}{3}+B-\frac{2}{3}=0-\frac{2}{3}$

  $B=-\frac{2}{3}$

Substitute 2/3 for A in 3^rd equation and solve for C

• By subtracting 2/3 on both sides, then
• Multiplying by -1 on both sides, and then

  $\frac{2}{3}-C=0$

  $\frac{2}{3}-C-\frac{2}{3}=0-\frac{2}{3}$

  $-C=-2/3$

  $-1\left(-C\right)=-1\left(-2/3\right)$

  $C=\frac{2}{3}$

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

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

Conclusion:

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