## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Section R.7 - Radical Expressions - R.7 Exercises: 84

#### Answer

$-14+11\sqrt{10}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(4\sqrt{5}+\sqrt{2})(3\sqrt{2}-\sqrt{5}) ,$ use the FOIL method. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 4\sqrt{5}(3\sqrt{2})+4\sqrt{5}(-\sqrt{5})+\sqrt{2}(3\sqrt{2})+\sqrt{2}(-\sqrt{5}) \\\\= 4(3)\sqrt{5(2)}-4\sqrt{5(5)}+3\sqrt{2(2)}-\sqrt{2(5)} \\\\= 12\sqrt{10}-4\sqrt{(5)^2}+3\sqrt{(2)^2}-\sqrt{10} \\\\= 12\sqrt{10}-4(5)+3(2)-\sqrt{10} \\\\= 12\sqrt{10}-20+6-\sqrt{10} \\\\= (-20+6)+(12\sqrt{10}-\sqrt{10}) \\\\= -14+11\sqrt{10} .\end{array}

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