Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 463: 60

Answer

$-10i$

Work Step by Step

Using $i^2=-1$, the expression $ (2-4i)(2-i) $ simplifies to \begin{array}{l} 2(2)+2(-i)-4i(2)-4i(-i) \text{(use the FOIL method)} \\= 4-2i-8i+4i^2 \\= 4-10i+4(-1) \\= -10i .\end{array}
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