College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.4 - Page 61: 47



Work Step by Step

$$(4x^2-1)^2$$ $$=(4x^2-1)(4x^2-1)$$ $$=(4x^2\times 4x^2)+(4x^2\times (-1))+(-1\times 4x^2)+(-1\times (-1))$$ $$=16x^4-4x^2-4x^2+1$$ $$=16x^4-8x^2+1$$
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