Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Test: 26


$x_{1}=1$ and $x_{2}=-\dfrac{2}{3}$

Work Step by Step

$(x-1)(3x^2-x-2)=0$ $(x-1)(3x^2-3x+2x-2)=0$ $(x-1)[3x(x-1)+2(x-1)]=0$ $(x-1)(x-1)(3x+2)=0$ $(x-1)^2(3x+2)=0$ Therefore the solutions are $x_{1}=1$ and $x_{2}=-\dfrac{2}{3}$
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