Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.4 - Multiplying Polynomials - Exercise Set - Page 289: 26

Answer

$12y^2-\dfrac{3}{2}y+\dfrac{1}{24}$

Work Step by Step

Using the FOIL Method, $(a+b)(c+d)=ac+ad+bc+bd$, the given expression simplifies to \begin{align*} & \left( 4y-\dfrac{1}{3} \right)\left( 3y-\dfrac{1}{8} \right) \\\\&= 12y^2-\dfrac{1}{2}y-y+\dfrac{1}{24} \\\\&= 12y^2-\dfrac{1}{2}y-\dfrac{2}{2}y+\dfrac{1}{24} \\\\&= 12y^2-\dfrac{3}{2}y+\dfrac{1}{24} .\end{align*}
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