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 - Test: Chapter 10: 28

Answer

$15-8i$

Work Step by Step

Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (4-i)^2 \\\\= (4)^2-2(4)(i)+(i)^2 \\\\= 16-8i+i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16-8i+i^2 \\\\= 16-8i+(-1) \\\\= 16-8i-1 \\\\= 15-8i .\end{array}
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