## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises - Page 475: 35

#### Answer

$4\sqrt{4y^2}-19\sqrt{2y}-5$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(\sqrt{2y}-5)(4\sqrt{2y}+1) ,$ use the special product on multiplying the sum and difference of like terms. $\bf{\text{Solution Details:}}$ Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \sqrt{2y}(4\sqrt{2y})+\sqrt{2y}(1)-5(4\sqrt{2y})-5(1) \\\\= 1(4)\sqrt{2y}(\sqrt{2y})+1(1)\sqrt{2y}-5(4)(\sqrt{2y})-5(1) \\\\= 4\sqrt{2y}(\sqrt{2y})+\sqrt{2y}-20\sqrt{2y}-5 .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4\sqrt{2y(2y)}+\sqrt{2y}-20\sqrt{2y}-5 \\\\= 4\sqrt{4y^2}+\sqrt{2y}-20\sqrt{2y}-5 .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 4\sqrt{4y^2}+(1-20)\sqrt{2y}-5 \\\\= 4\sqrt{4y^2}-19\sqrt{2y}-5 .\end{array}

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