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: 70

Answer

$16x^2-1$

Work Step by Step

Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the expression, $ (4x+1)(4x-1) $, is equivalent to \begin{array}{l} (4x)^2-(1)^2 \\\\= 16x^2-1 .\end{array}
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