Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Section 5.6 - Special Products - Practice: 16

Answer

$4a^{2}-1$

Work Step by Step

Step 1: $(2a−1)(2a+1)$ Step 2: Using a special product to multiply, $(2a−1)(2a+1)=(2a)^{2}-(1)^{2}$ Step 3: $(2a)^{2}-(1)^{2}=4a^{2}-1$
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