## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 4 - Polynomials - Test: Chapter 4: 30

#### Answer

$x^2+\dfrac{2}{3}x+\dfrac{1}{9}$

#### Work Step by Step

Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the given expression, $\left( x-\dfrac{1}{3} \right)^2 ,$ is \begin{array}{l}\require{cancel} (x)^2+2(x)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)^2 \\\\= x^2+\dfrac{2}{3}x+\dfrac{1}{9} .\end{array}

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