Intermediate Algebra (12th Edition)

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

Chapter 4 - Chapters R-4 - Cumulative Review Exercises: 34



Work Step by Step

Dividing each of the terms of the numerator by the denominator, the given expression, $ \dfrac{16x^3y^5-8x^2y^2+4}{4x^2y} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{16x^3y^5}{4x^2y}-\dfrac{8x^2y^2}{4x^2y}+\dfrac{4}{4x^2y} .\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} 4x^{3-2}y^{5-1}-2x^{2-2}y^{2-1}+x^{-2}y^{-1} \\\\= 4x^{1}y^{4}-2x^{0}y^{1}+x^{-2}y^{-1} \\\\= 4xy^{4}-2y+x^{-2}y^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4xy^{4}-2y+\dfrac{1}{x^{2}y^1} \\\\= 4xy^{4}-2y+\dfrac{1}{x^{2}y} .\end{array}
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