Intermediate Algebra (6th Edition)

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

Chapter 5 - Review - Page 331: 138


$$x=0$$ $$x=\frac{1}{4}$$ $$x=-\frac{3}{2}$$

Work Step by Step

We have $$8x^3+10x^2=3x$$ Sending all the term to the left side of the equation $$8x^3+10x^2-3x=0$$ Factorizing $$x(8x^2+10x-3)=0$$ $$x(4x-1)(2x+3)=0$$ Solving each parenthesis 1)$$x=0$$ 2)$$4x-1=0$$ $$4x=1$$ $$x=\frac{1}{4}$$ 3)$$2x+3=0$$ $$2x=-3$$ $$x=-\frac{3}{2}$$
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