Intermediate Algebra (6th Edition)

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

Chapter 5 - Test - Page 332: 27


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

Work Step by Step

The factored form of the given expression, $ 2x^3+5x^2=8x+20 $, is \begin{array}{l} 2x^3+5x^2-8x-20=0 \\\\= (2x^3+5x^2)-(8x+20)=0 \\\\= x^2(2x+5)-4(2x+5)=0 \\\\= (2x+5)(x^2-4)=0 \\\\= (2x+5)(x+2)(x-2)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} 2x+5=0 \\ 2x=0-5 \\ 2x=-5 \\ x=-\dfrac{5}{2} ,\text{ OR}\\\\ x+2=0 \\ x=0-2 \\ x=-2 ,\text{ OR}\\\\ x-2=0 \\ x=0+2 \\ x=2 .\end{array} Hence, $ x=\left\{ -\dfrac{5}{2},-2,2 \right\} $.
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