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: 43


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

Work Step by Step

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