Q. 52 - 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.

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

Step-by-Step Solution

Answer

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

Step 1. Given data

The given expression is 

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

Step 2. Division

Division Algorithm 

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

Step 3. Partial fraction

partial fraction decomposition of the proper fraction 

$$\begin{gathered} \frac{6x-3}{x^2-2x+1}=\frac{6x-3}{{\left(x-1\right)}^2} \\ \frac{6x-3}{{\left(x-1\right)}^2}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2} \\ 6x-3=A(x-1)+B \\ 6(x-1)+3=A(x-1)+B \\ So A=6 \\ B=3 \end{gathered}$$

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