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: 57

Answer

$18+13i$

Work Step by Step

Using $i^2=-1$, the expression $ (4+i)(5+2i) $ simplifies to \begin{array}{l} 4(5)+4(2i)+i(5)+i(2i) \\= 20+8i+5i+2i^2 \\= 20+13i+2(-1) \\= 18+13i .\end{array}
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