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: 58

Answer

$2+14i$

Work Step by Step

Using $i^2=-1$, the expression $ (3+i)(2+4i) $ simplifies to \begin{array}{l} 3(2)+3(4i)+i(2)+i(4i) \\= 6+12i+2i+4i^2 \\= 6+14i+4(-1) \\= 2+14i .\end{array}
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