Q. 54 - Precalculus Enhanced with Graphing Utilities

Question

In Problems, use the division algorithm to rewrite each improper fraction as the sum of a quotient and proper fraction. Find the partial fraction decomposition of the proper fraction. Finally, express the improper fraction as the sum of a quotient and the partial fraction decomposition.

$\frac{x^5-x^3+x^2+1}{x^4+6x^2+9}$

Step-by-Step Solution

Answer

Expression as the sum of a quotient and the partial fraction decomposition is $\frac{x^5-x^3+x^2+1}{x^4+6x^2+9}=x+\frac{-7x+1}{\left(x^2+3\right)}+\frac{12x-2}{{\left(x^2+3\right)}^2}$

Step 1. Given data

The given expression is 

$\frac{x^5-x^3+x^2+1}{x^4+6x^2+9}$

Step 2. Division

Division Algorithm 

So $\frac{x^5-x^3+x^2+1}{x^4+6x^2+9}=x+\frac{-7x^3+x^2-9x+1}{x^4+6x^2+9}$

Step 3. Partial fraction

partial fraction decomposition of the proper fraction 

$$\begin{gathered} \frac{-7x^3+x^2-9x+1}{x^4+6x^2+9}=\frac{-7x^3+x^2-9x+1}{{\left(x^2+3\right)}^2} \\ \frac{-7x^3+x^2-9x+1}{{\left(x^2+3\right)}^2}=\frac{Ax+B}{\left(x^2+3\right)}+\frac{Cx+D}{{\left(x^2+3\right)}^2} \\ -7x^3+x^2-9x+1=\left(Ax+B\right)\left(x^2+3\right)+\left(Cx+D\right) \\ \left(-7x+1\right)\left(x^2+3\right)+\left(12x-2\right)=\left(Ax+B\right)\left(x^2+3\right)+\left(Cx+D\right) \\ So A=-7, B=1, C=12, \& D=-2 \end{gathered}$$

so $\frac{x^5-x^3+x^2+1}{x^4+6x^2+9}=x+\frac{-7x+1}{\left(x^2+3\right)}+\frac{12x-2}{{\left(x^2+3\right)}^2}$