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

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