College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding: 86

Answer

$x=-\frac{4}{3}$ or $x=3$ or $x=-3$

Work Step by Step

We solve by factoring with grouping: $3x^{3}+4x^{2}=27x+36$ $3x^{3}+4x^{2}-27x-36=0$ $x^{2}(3x+4)-9(3x+4)=0$ $(3x+4)(x^{2}-9)=0$ $(3x+4)(x-3)(x+3)=0$ $3x+4=0$ or $x-3=0$ or $x+3=0$ $3x=-4$ or $x=3$ or $x=-3$ $x=-\frac{4}{3}$ or $x=3$ or $x=-3$
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