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 - Page 289: 60



Work Step by Step

$(4x^{2}+4x-4)^{2} $ $=[4(x^{2}+x-1)]^{2} $ $=4^{2}(x^{2}+x-1)^{2} $ $=16(x^{2}+x-1)^{2} $ $=16(x^{2}+(x-1))^{2} $ Using the formula $(a+b)^{2}= a^{2}+2ab+b^{2}$ Let $a$ be $x^{2}$ and $b$ be $(x-1)$ $(x^{2})^{2}+2(x^{2})(x-1)+(x-1)^{2}$ $=16[x^{4}+2x^{2}(x-1)+(x-1)^{2}]$ [Using $(a-b)^{2}= a^{2}-2ab+b^{2}$ $(x-1)^{2}= x^{2}-2x+1 $] $=16[x^{4}+2x^{3}-2x^{2}+x^{2}-2x+1]$ Combine like terms. $= 16[x^{4}+2x^{3}-x^{2}-2x+1]$
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