Answer
$x_{1}=1$ 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 $x_{2}=-\dfrac{2}{3}$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 6 - Test - Page 478: 26
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