Chapter 10, Problem 88RE

To determine

To find: The Partial fraction decomposition of the rational function

Answer to Problem 88RE

The partial decomposition is $\frac{x+6}{x^3-2x^2+4x-8}$ is $\frac{1}{x-2}-\frac{x+1}{x^2+4}$

Explanation of Solution

Given:

  $\frac{x+6}{x^3-2x^2+4x-8}$

Calculation:

Consider the rational function $r(x)=\frac{x+6}{x^3-2x^2+4x-8}$

To find the partial function decomposition write the denominator as the product of factors

  $x^3-2x^2+4x-8=x^2(x-2)+4(x-2)$

  $=(x^2+4)·(x-2)$

  $\frac{x+6}{x^3-2x^2+4x-8}=\frac{A}{x-2}+\frac{Bx+C}{x^2+4}\cdots \cdots (1)$

  $x+6=\left(\frac{A}{x-2}+\frac{Bx+C}{x^2+4}\right)\left(x^3-2x^2+4x-8\right)$

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

  $\begin{array}{l}x+6=A{x}^2+4A+B{x}^2-2Bx+Cx-2C\\ x+6=\left(A+B\right)·x^2+(-2B+C)·x+(4A-2C)\end{array}$

Compare the coefficients of $x^2$ , $x$ , constant term on both sides

  $A+B=0$ The x term has a coefficient of $1$ , so $-2B+C=1$ .

Finally the last term is $6$ , So

  $4A-2C=6$

This gives the system

  $\begin{array}{l}A+B=0\\ -2B+C=1\\ 4A-2C=6\end{array}$

Multiply (1) with 2 and add to the (2)

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

  $2A+C=1$

Call this asequation (4)

Multiply (4) equation with 2 and add to theequation (3)

  $\begin{array}{l}4A+2C=2\\ 4A-2C=6\end{array}$

  $8A=8$

  $8$

Use the value of Ain theequation (1), $1+B=0$

So $B=-1$

Also use $A=1$ in the 3^rdequation, $-2C=6-4(1)$

  $C=-1$

Replace A,B,C values in (1), the required partial decomposition is

  $\begin{array}{l}\frac{x+6}{x^3-2x^2+4x-8}=\frac{1}{x-2}+\frac{(-1)x+(-1)}{x^2+4}\\ =\frac{1}{x-2}-\frac{x+1}{x^2+4}\end{array}$

Conclusion:

Hence,the partial decomposition of the rational function $\frac{x+6}{x^3-2x^2+4x-8}$ is $\frac{1}{x-2}-\frac{x+1}{x^2+4}$