Chapter 7.4, Problem 33E

To determine

To write : the partial fraction decomposition of the rational expression

Answer to Problem 33E

  $\frac{x}{16x^4-1}=\boxed{\frac{1}{8\left(2x-1\right)}+\frac{1}{8\left(2x+1\right)}-\frac{x}{2\left(4x^2+1\right)}}$

Explanation of Solution

Given information: $\frac{x}{16x^4-1}$

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 expression $16x^4-1$

  $\frac{x}{16x^4-1}=\frac{x}{{\left(4x^2\right)}^2-1^2}$

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

  $\frac{x}{{\left(4x^2\right)}^2-1^2}=\frac{x}{\left(4x^2-1\right)\left(4x^2+1\right)}$

Rewrite the expression $4x^2-1$

  $\frac{x}{\left(4x^2-1\right)\left(4x^2+1\right)}=\frac{x}{\left({\left(2x\right)}^2-1^2\right)\left(4x^2+1\right)}$

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

  $\frac{x}{\left({\left(2x\right)}^2-1^2\right)\left(4x^2+1\right)}=\frac{x}{\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}$

Write one partial fraction for each of those factors

  $\frac{x}{\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}=\frac{A}{2x+1}+\frac{B}{2x-1}+\frac{Cx+D}{4x^2+1}$

Multiply through by the LCD $\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)$ so we no longer have fractions

  $\begin{array}{l}\frac{x\cdot \left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}{\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}=\frac{A\cdot \left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}{2x+1}+\frac{B\cdot \left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}{2x-1}\\ +\frac{\left(Cx+D\right)\cdot \left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}{4x^2+1}\end{array}$

Cancel common factors throughout the equation

Expand the expression in the right side of the equation

  $x=A\left(8x^3+2x-4x^2-1\right)+B\left(8x^3+2x+4x^2+1\right)+4C{x}^3-Cx+4D{x}^2-D$

Distribute in right side of the equation

  $x=8A{x}^3+2Ax-4A{x}^2-A+8B{x}^3+2Bx+4B{x}^2+B+4C{x}^3-Cx+4D{x}^2-D$

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

  $x=8A{x}^3+8B{x}^3+4C{x}^3-4A{x}^2+4B{x}^2+4D{x}^2+2Ax+2Bx-Cx-A+B-D$

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

  $0x^3+0x^2+1x+0=\left(8A+8B+4C\right)x^3+\left(-4A+4B+4D\right)+\left(2A+2B-C\right)+\left(-A+B-D\right)$

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

  $8A+8B+4C=0$

  $-4A+4B+4D=0$

  $2A+2B-C=1$

  $-A+B-D=0$

Label the equation we got

  $8A+8B+4C=0$  ▶ 1^st equation

  $-4A+4B+4D=0$  ▶ 2^nd equation

  $2A+2B-C=1$  ▶ 3^rd equation

  $-A+B-D=0$  ▶ 4^th equation

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

  $\begin{array}{l}-8A-8B+4C=-4\\ \underset{\_}{8A+8B+4C=0}\\ 8C=-4\end{array}$

Solve the equation $8C=-4$

• By dividing 8 on both sides, then
• Simplifying fraction on both sides of the equation

  $8C/8=-4/8$

  $C=-1/2$

In order to eliminate variable C multiply 4 with the 3^rd equation and add the resulting equation to the 1^st equation to get the 5^th equation

  $\begin{array}{l}8A+8B-4C=4\\ \underset{\_}{8A+8B+4C=0}\\ 16A+16B=4\end{array}$

Label the equation

  $16A+16B=4$  ▶ 5^th equation

In order to eliminate variable D multiply 4 with the 4^th equation and add the resulting equation to the 2^nd equation to get the 6^th equation

  $\begin{array}{l}-4A+4B-4D=0\\ \underset{\_}{-4A+4B+4D=0}\\ -8A+8B=0\end{array}$

Label the equation

  $-8A+8B=0$  ▶ 6^th equation

In order to eliminate variable A multiply 2 with 6^th equation and add the resulting equation to the 5^th equation

  $\begin{array}{l}-16A+16B=0\\ \underset{\_}{16A+16B=4}\\ 32B=4\end{array}$

Solve the equation $32B=4$

• By dividing 32 on both sides, then
• Simplifying fraction on both sides of the equation

  $\frac{32B}{32}=\frac{4}{32}$

  $B=1/8$

Substitute 1/8 for Band $-1/2$ for C in the 3^rd equation and Solve the equation for A

• Simplify fraction in left side of the equation, then
• Adding the fractions, then
• By subtracting 3/4 on both sides, and then
• Dividing by 2 on both sides
• Flip the second fraction and multiply with the first in the right side of the equation

  $2A+2\left(1/8\right)-\left(-1/2\right)=1$

  $2A+1/4+1/2=1$

  $2A+1/4+2/4=1$

  $2A+3/4=1$

  $2A+3/4-3/4=1-3/4$

  $2A=4/4-3/4$

  $2A=1/4$

  $2A÷2=1/4÷2$

  $A=1/4\cdot 1/2$

  $A=1/8$

Substitute 1/8 for A and 1/8 for B in the 4^th equation and solve the equation for D

• By adding fractions in the left side of the equation, and then
• Multiply -1 on both sides of the equation

  $-1/8+1/8-D=0$

  $-D=0$

  $-1\cdot \left(-D\right)=-1\cdot \left(0\right)$

  $D=0$

Substitute the value of A, B, C, D in the equation $\frac{x}{\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}=\frac{A}{2x+1}+\frac{B}{2x-1}+\frac{Cx+D}{4x^2+1}$

  $\frac{x}{\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)}=\frac{1/8}{2x+1}+\frac{1/8}{2x-1}+\frac{1/2x}{4x^2+1}$

Conclusion:

The partial fraction decomposition of the given rational expressionis: $\boxed{\frac{1}{8\left(2x-1\right)}+\frac{1}{8\left(2x+1\right)}-\frac{x}{2\left(4x^2+1\right)}}$