Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.4 - Multiplying Polynomials - Exercise Set - Page 289: 82


$24a^{3} +28a^{2}b-6a-7b$

Work Step by Step

$(2a+1)(2a-1)(6a+7b)$ Multiplying $(2a+1)(2a-1),$ it is in the form of $(a+b)(a-b)$ $(a+b)(a-b)= (a^{2}-b^{2})$. $(2a+1)(2a-1)=(2a)^{2}-1^{2} = 4a^{2} - 1$ $(2a+1)(2a-1)(6a+7b) = (4a^{2} - 1)(6a+7b)$ $=24a^{3}+28a^{2}b-6a-7b$
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