## College Algebra (10th Edition)

Published by Pearson

# Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding: 85

#### Answer

$x=\frac{1}{2}$ or $x=2$ or $x=-2$

#### Work Step by Step

We solve by factoring with grouping: $2x^{3}+4=x^{2}+8x$ $2x^{3}+4-x^{2}-8x=0$ $2x^{3}-x^{2}-8x+4=0$ $x^{2}(2x-1)-4(2x-1)=0$ $(2x-1)(x^{2}-4)=0$ $(2x-1)(x-2)(x+2)=0$ $2x-1=0$ or $x-2=0$ or $x+2=0$ $2x=1$ or $x=2$ or $x=-2$ $x=\frac{1}{2}$ or $x=2$ or $x=-2$

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