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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - Review Exercises: Chapter 4: 58

Answer

$a^2+\dfrac{1}{6}a-\dfrac{1}{3}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of 2 binomials, the product of the given expression, $ \left( a-\dfrac{1}{2} \right)\left( a+\dfrac{2}{3} \right) ,$ is \begin{array}{l}\require{cancel} a(a)+a\left(\dfrac{2}{3}\right)-\dfrac{1}{2}(a)-\dfrac{1}{2}\left(\dfrac{2}{3}\right) \\\\= a^2+\dfrac{2}{3}a-\dfrac{1}{2}a-\dfrac{1}{3} \\\\= a^2+\dfrac{4}{6}a-\dfrac{3}{6}a-\dfrac{1}{3} \\\\= a^2+\dfrac{1}{6}a-\dfrac{1}{3} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.