Intermediate Algebra (6th Edition)

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

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 9



Work Step by Step

Since the first 3 terms of the given expression, $ x^2-8x+16-y^2 $, form a perfect square trinomial, then, \begin{array}{l} (x^2-8x+16)-y^2 \\\\= (x-4)^2-y^2 .\end{array} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{array}{l} (x-4)^2-y^2 \\\\= [(x-4)+y][(x-4)-y] \\\\= (x-4+y)(x-4-y) .\end{array}
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