Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Chapter R-9 - Cumulative Review Exercises - Page 642: 18

Answer

$16k^2-24k+9$

Work Step by Step

Using $(a-b)^2=a^2-2ab+b^2$, the given expression, $ (4k-3)^2 ,$ is equivalent to \begin{align*} & (4k)^2-2(4k)(3)+(3)^2 \\&= 16k^2-24k+9 .\end{align*} Hence, the expression $ (4k-3)^2 $ is equivalent to $ 16k^2-24k+9 $.
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