Intermediate Algebra (12th Edition)

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

Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 491: 53



Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (4+3i)(1-2i) ,$ use the FOIL method. Then use $i^2=-1$ and combine like terms. $\bf{\text{Solution Details:}}$ 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(1)+4(-2i)+3i(1)+3i(-2i) \\\\= 4-8i+3i-6i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4-8i+3i-6(-1) \\\\= 4-8i+3i+6 .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} (4+6)+(-8i+3i) \\\\= 10-5i .\end{array}
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