College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Summary, Review, and Test - Review Exercises - Page 204: 56

Answer

$-8(12+5i)$

Work Step by Step

$(-2+\sqrt -100)^{2}$ Using $\sqrt -b = i\sqrt b$ $(-2+\sqrt -100)^{2} = (-2+i\sqrt 100)^{2}$ $=(-2+10i)^{2}$ Using $(a+b)^{2}=(a)^{2}+2(a)(b)+(b)^{2}$ $(-2+10i)^{2}=(-2)^{2}+2(-2)(10i)+(10i)^{2}$ $=4-40i+100i^{2}$ [Substituting $i^{2}=-1$] $=4-40i+100(-1)$ $=4-40i-100$ $=-40i-96$ $=-8(5i+12)$ In standard form, $-8(12+5i)$
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