## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 50

#### Answer

$16+10\sqrt{10}$

#### Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$, or the product of 2 binomials, and the properties of radicals, the given expression, $(4\sqrt{5}-3\sqrt{2})(2\sqrt{5}+4\sqrt{2}) ,$ is equivalent to \begin{array}{l}\require{cancel} (4\sqrt{5})(2\sqrt{5})+(4\sqrt{5})(4\sqrt{2})-(3\sqrt{2})(2\sqrt{5})-(3\sqrt{2})(4\sqrt{2}) \\\\= 8\sqrt{25}+16\sqrt{10}-6\sqrt{10}-12\sqrt{4} \\\\= 8\cdot5+16\sqrt{10}-6\sqrt{10}-12\cdot2 \\\\= 40+16\sqrt{10}-6\sqrt{10}-24 \\\\= (40-24)+(16\sqrt{10}-6\sqrt{10}) \\\\= 16+10\sqrt{10} .\end{array}

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