College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises: 27

Answer

$-30-40i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ -5i(3-i)^2 ,$ use the special product on squaring binomials and the Distributive Property. Then use the equivalence $i^2=-1.$ Finally, combine combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ 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} -5i[(3)^2-2(3)(i)+(i)^2] \\\\= -5i[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} -5i(9)-5i(-6i)-5i(i^2) \\\\= -45i+30i^2-5i(i^2) .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -45i+30(-1)-5i(-1) \\\\= -45i-30+5i .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} -30+(-45i+5i) \\\\= -30-40i .\end{array}
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