Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 44


$r=\left\{ -3,0,\dfrac{4}{5} \right\}$

Work Step by Step

Equating each factor to zero, then the solution to the given equation, $ 2r(r+3)(5r-4)=0 $, is \begin{array}{l} 2r=0 \\ r=\dfrac{0}{2} \\ r=0 ,\text{ OR}\\\\ r+3=0 \\ r=0-3 \\ r=-3 ,\text{ OR}\\\\ 5r-4=0 \\ 5r=0+4 \\ 5r=4 \\ r=\dfrac{4}{5} \end{array} Hence, $ r=\left\{ -3,0,\dfrac{4}{5} \right\} $.
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