Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.5 Partial Fraction Decomposition - 10.5 Assess Your Understanding - Page 788: 10


$\text{improper}$; $3x+\dfrac{x^2-24x-2}{x^{3}+8}$

Work Step by Step

A rational expression $\dfrac{A(x)}{B(x)}$ is said to be proper if the degree of polynomial in numerator is less than the degree of its denominator.If it does not happen , then it is said to be improper rational polynomial. Here, the degree of the numerator $A(x)= 3x^4+x^2-21$ is $4$ and the degree of the denominator ; $B(x)=x^3+8$ is $3$. We see that the given rational expression is improper. To make the rational expression proper, we will solve as: $ \dfrac{3x^4+x^2-2}{x^{3}+8}=\dfrac{3x^{4}+24x+x^2-24x-2}{x^{3}+8} \\=\dfrac{(3x^{4}+24x)+(x^2-24x-2)}{x^{3}+8} \\=\dfrac{3x(x^{3}+8)+(x^2-24x-2)}{x^{3}+8} \\=\dfrac{3x(x^{3}+8)}{x^{3}+8}+\dfrac{x^2-24x-2}{x^{3}+8} \\=3x+\dfrac{x^2-24x-2}{x^{3}+8}$
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