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 - Page 368: 59



Work Step by Step

Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the square of the given expression, $\left( 4y-\dfrac{1}{4} \right)^2,$ is \begin{array}{l}\require{cancel} (4y)^2-2(4y)\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{4} \right)^2 \\\\= 16y^2-2y+\dfrac{1}{16} .\end{array}
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