## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Review - Page 330: 68

#### Answer

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

#### Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the FOIL Method, then, \begin{array}{l} \left( x-\dfrac{1}{3} \right)\left( x+\dfrac{2}{3} \right) \\\\= x^2+\dfrac{2}{3}x-\dfrac{1}{3}x-\dfrac{2}{9} \\\\= x^2+\dfrac{1}{3}x-\dfrac{2}{9} .\end{array}

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