Q. 51 - 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.

$\frac{x^4-5x^2+x-4}{x^2+4x+4}$

Step-by-Step Solution

Answer

Expression $\frac{x^4-5x^2+x-4}{x^2+4x+4}$as the sum of a quotient and the partial fraction decomposition is$\frac{x^4-5x^2+x-4}{x^2+4x+4}=x^2-4x+7-\frac{11}{x+2}-\frac{10}{{\left(x+2\right)}^2}$

Step 1. Given data

The given expression is

$\frac{x^4-5x^2+x-4}{x^2+4x+4}$

Step 2. Division

Division Algorithm

So $\frac{x^4-5x^2+x-4}{x^2+4x+4}=x^2-4x+7-\frac{11x+32}{x^2+4x+4}$

Step 3. Partial fraction

partial fraction decomposition of the proper fraction

$$\begin{gathered} \frac{11x+32}{x^2+4x+4}=\frac{11x+32}{{\left(x+2\right)}^2} \\ \frac{11x+32}{{\left(x+2\right)}^2}=\frac{A}{\left(x+2\right)}+\frac{B}{{\left(x+2\right)}^2} \\ 11x+32=A\left(x+2\right)+B \\ 11(x+2)+10=A\left(x+2\right)+B \\ So A=11 \\ B=10 \end{gathered}$$

so$\frac{x^4-5x^2+x-4}{x^2+4x+4}=x^2-4x+7-\frac{11}{x+2}-\frac{10}{{\left(x+2\right)}^2}$