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: 76

Answer

$4a^4+12a^2+9$

Work Step by Step

Using $(a+b)^2=a^2+2ab+b^2$ or the Square of a Binomial, the expression, $ [(2a^2+4)-1]^2 $, is equivalent to \begin{array}{l} (2a^2+4)^2+2(2a^2+4)(-1)+(-1)^2 \\\\= (2a^2)^2+2(2a^2)(4)+(4)^2+2(2a^2+4)(-1)+(-1)^2 \\\\= 4a^4+16a^2+16-4a^2-8+1 \\\\= 4a^4+(16a^2-4a^2)+(16-8+1) \\\\= 4a^4+12a^2+9 .\end{array}
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