Chapter 7.4, Problem 34E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 34E

  $\frac{3}{x^4+x}=\boxed{\frac{3}{x}-\frac{1}{x+1}+\frac{1-2x}{x^2-x+1}}$

Explanation of Solution

Given information: $\frac{3}{x^4+x}$

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 x from the denominator of the rational expression

  $\frac{3}{x^4+x}=\frac{3}{x\left(x^3+1\right)}$

Rewrite the denominator of the rational expression

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

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

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

Write one partial fraction for each of those factors

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

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

  $\begin{array}{l}\frac{3\cdot x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x+1\right)\left(x^2-x+1\right)}=\frac{A\cdot x\left(x+1\right)\left(x^2-x+1\right)}{x}+\frac{B\cdot x\left(x+1\right)\left(x^2-x+1\right)}{x+1}\\ +\frac{\left(Cx+D\right)\cdot x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}\end{array}$

Cancel common factors throughout the equation

  $3=A\left(x+1\right)\left(x^2-x+1\right)+Bx\left(x^2-x+1\right)+\left(Cx+D\right)\cdot x\left(x+1\right)$

Expand the right side of the equation

  $3=A{x}^3-A{x}^2+Ax+A{x}^2-Ax+A+B{x}^3-B{x}^2+Bx+C{x}^3+C{x}^2+D{x}^2+Dx$

Group and combine $x^3,x^2,x,\&$ constant terms in right side of the equation

  $3=A{x}^3+B{x}^3+C{x}^3-B{x}^2+C{x}^2+D{x}^2+Bx+Dx+A$

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

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

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

  $A+B+C=0$

  $-B+C+D=0$

  $B+D=0$

  $A=3$

Label the equation we got

  $A+B+C=0$  ▶ 1^st equation

  $-B+C+D=0$  ▶ 2^nd equation

  $B+D=0$  ▶ 3^rd equation

In order to eliminate variable C, multiply -1 to the first equation and add the resulting equation to 2^nd equation

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

Label the resulting equation as 4^th equation

  $-A-2B+D=0$  ▶ 4^th equation

In order to eliminate the variable D, multiply -1 to the 4^th equation and add the resulting equation to 3^rd equation

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

Substitute 3 for A in the equation $A+3B=0$ &Solve the equation $3+3B=0$ for B

• By subtracting 3 on both sides, then
• Bydividing 3 on both sides, and
• Simplifying fraction on both sides

  $3+3B-3=0-3$

  $3B=-3$

  $\frac{3B}{3}=\frac{-3}{3}$

  $B=-1$

Substitute 3 for A and -1 for B in 1^st equation and solve for C

• By subtracting2 on both sides

  $3+-1+C=0$

  $2+C-2=0-2$

  $C=-2$

Substitute -1 for Bin 3^rd equation and solve for D

• By adding 1 on both sides, and then

  $-1+D=0$

  $-1+D+1=0+1$

  $D=1$

Substitute the value of A, B, C and D in the equation $\frac{3}{x\left(x+1\right)\left(x^2-x+1\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{x^2-x+1}$

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

Conclusion:

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