## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 491: 58

#### Answer

$-18+24i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $3i(-3-i)^2 ,$ 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} 3i[(-3)^2+2(-3)(-i)+(-i)^2] \\\\= 3i[9+6i+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} 3i(9)+3i(6i)+3i(i^2) \\\\= 27i+18i^2+3i(i^2) .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 27i+18(-1)+3i(-1) \\\\= 27i-18-3i .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} -18+(27i-3i) \\\\= -18+24i .\end{array}

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