Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Section 5.3 - Special Products - Exercise Set: 32

Answer

$16-s^2$

Work Step by Step

Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the product of the given expression, $ (4+s)(4-s) ,$ is \begin{array}{l}\require{cancel} 4(4)-s(s) \\\\= 16-s^2 .\end{array}
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