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: 60


$x=\left\{ -\dfrac{13}{5},0,2 \right\}$

Work Step by Step

The factored form of the given expression, $ x^2(5x+3)=26x $, is \begin{array}{l} 5x^3+3x^2-26x=0 \\\\ x(5x^2+3x-26)=0 \\\\ x(5x+13)(x-2)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} x=0 ,\text{ OR}\\\\ 5x+13=0 \\ 5x=0-13 \\ 5x=-13 \\ x=-\dfrac{13}{5} ,\text{ OR}\\\\ x-2=0 \\ x=0+2 \\ x=2 \end{array} Hence, $ x=\left\{ -\dfrac{13}{5},0,2 \right\} $.
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