Intermediate Algebra (6th Edition)

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

Chapter 11 - Section 11.5 - The Binomial Theorem - Exercise Set - Page 664: 23

Answer

$1024 a^5 + 1280 a^4 b + 640 a^3 b^2 + 160 a^2 b^3 + 20 a b^4 + b^5$

Work Step by Step

Using the Binomial Formula, the expression $ (4a+b)^5 $ expands to \begin{array}{l} (4a)^5b^0+ \dfrac{5}{1!}(4a)^4b^1+ \dfrac{5\cdot4}{2!}(4a)^3b^2+ \dfrac{5\cdot4\cdot3}{3!}(4a)^2b^3+\\ \dfrac{5\cdot4\cdot3\cdot2}{4!}(4a)^1b^4+ \dfrac{5\cdot4\cdot3\cdot2\cdot1}{5!}(4a)^0b^5 \\\\= 1024 a^5 + 1280 a^4 b + 640 a^3 b^2 + 160 a^2 b^3 + 20 a b^4 + b^5 \end{array}
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.