Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.4 - Complex Numbers - Problem Set 11.4: 42

Answer

$14+32i$

Work Step by Step

We multiply each term of the first complex number with the second complex number and then simplify: $(4+2i)(6+5i)$ =$4(6+5i)+2i(6+5i)$ =$24+20i+12i+10i^{2}$ =$24+32i+10(-1)$ [We substitute -1 in place of $i^{2}$ as $i^{2}=-1$] =$24-10+32i$ =$14+32i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.