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

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 686: 50

#### Answer

$5+14i$

#### 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, $(4+i)(2+3i) ,$ simplifies to \begin{array}{l}\require{cancel} 4(2)+4(3i)+i(2)+i(3i) \\\\= 8+12i+2i+3i^2 \\\\= 8+14i+3i^2 \\\\= 8+14i+3(-1) \\\\= 8+14i-3 \\\\= 5+14i .\end{array}

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