Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Review Exercises - Page 578: 93



Work Step by Step

Solve.$(7-5i)(2+3i)$ we have, $=7(2+3i)-5i(2+3i)$ Need to drop the parentheses and multiply the terms. $=14+21i-10i-15i^2$ Since, $i^2=-1$ Thus, $=14+11i-15(-1)$ or, $=29+11i$
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