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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 694: 57



Work Step by Step

Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (6-3i)(2-i) \\\\= 6(2)+6(-i)-3i(2)-3i(-i) \\\\= 12-6i-6i+3i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 12-6i-6i+3i^2 \\\\= 12-6i-6i+3(-1) \\\\= 12-6i-6i-3 \\\\= (12-3)+(-6i-6i) \\\\= 9-12i .\end{array}
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