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Answers without the blur. Sign up and see all textbooks for free! Q. 53

Expert-verified Found in: Page 765 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # 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.$\frac{{x}^{5}+{x}^{4}-{x}^{2}+2}{{x}^{4}-2{x}^{2}+1}$

Expression $\frac{{x}^{5}+{x}^{4}-{x}^{2}+2}{{x}^{4}-2{x}^{2}+1}$as the sum of a quotient and the partial fraction decomposition is $\frac{{x}^{5}+{x}^{4}-{x}^{2}+2}{{x}^{4}-2{x}^{2}+1}=x+1+\frac{1.25}{\left(x-1\right)}+\frac{0.25}{{\left(x-1\right)}^{2}}+\frac{0.75}{\left(x+1\right)}+\frac{0.25}{{\left(x+1\right)}^{2}}$

See the step by step solution

## Step 1. Given data

The given expression is

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

## Step 2. Division

Division Algorithm $\frac{{x}^{5}+{x}^{4}-{x}^{2}+2}{{x}^{4}-2{x}^{2}+1}=x+1+\frac{2{x}^{3}+{x}^{2}-x+1}{{x}^{4}-2{x}^{2}+1}$

## Step 3. Partial fraction

partial fraction decomposition of the proper fraction

$\frac{2{x}^{3}+{x}^{2}-x+1}{{x}^{4}-2{x}^{2}+1}=\frac{2{x}^{3}+{x}^{2}-x+1}{{\left(x-1\right)}^{2}{\left(x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{2{x}^{3}+{x}^{2}-x+1}{{\left(x-1\right)}^{2}{\left(x+1\right)}^{2}}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^{2}}+\frac{C}{\left(x+1\right)}+\frac{D}{{\left(x+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}2{x}^{3}+{x}^{2}-x+1=A\left(x-1\right){\left(x+1\right)}^{2}+B{\left(x+1\right)}^{2}+C\left(x+1\right){\left(x-1\right)}^{2}+D{\left(x-1\right)}^{2}$

Solving A,B,C, and D

$A=1.25,B=0.25,C=0.75,&D=0.25$

So $\frac{{x}^{5}+{x}^{4}-{x}^{2}+2}{{x}^{4}-2{x}^{2}+1}=x+1+\frac{1.25}{\left(x-1\right)}+\frac{0.25}{{\left(x-1\right)}^{2}}+\frac{0.75}{\left(x+1\right)}+\frac{0.25}{{\left(x+1\right)}^{2}}$ ### Want to see more solutions like these? 