Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Review Exercises - Page 318: 63



Work Step by Step

Dividing each of the terms of the numerator by the denominator, the given expression, $ \dfrac{4y^3-12y^2+5y}{4y} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4y^3}{4y}-\dfrac{12y^2}{4y}+\dfrac{5y}{4y} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} y^{3-1}-3y^{2-1}+\dfrac{5\cancel{y}}{4\cancel{y}} \\\\= y^{2}-3y^{1}+\dfrac{5}{4} \\\\= y^{2}-3y+\dfrac{5}{4} .\end{array}
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