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: 58

Answer

$9x^2+2x+\dfrac{1}{9}$

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, $\left( 3x+\dfrac{1}{3} \right)^2,$ is \begin{array}{l}\require{cancel} (3x)^2+2(3x)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)^2 \\\\= 9x^2+2x+\dfrac{1}{9} .\end{array}
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