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: 57

Answer

$-16+30i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ 2i(-4-i)^2 ,$ use the special product on squaring binomials and the Distributive Property. Then use $i^2=-1$ and combine like terms. $\bf{\text{Solution Details:}}$ 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 expression above is equivalent to \begin{array}{l}\require{cancel} 2i[(-4)^2+2(-4)(-i)+(-i)^2] \\\\= 2i[16+8i+i^2] .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2i(16)+2i(8i)+2i(i^2) \\\\= 32i+16i^2+2i(i^2) .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 32i+16(-1)+2i(-1) \\\\= 32i-16-2i .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} -16+(32i-2i) \\\\= -16+30i .\end{array}
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