College Algebra 7th Edition

Published by Brooks Cole

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 138: 24

Answer

$x=\frac{1}{2}$ or $x=\pm\sqrt{\frac{1}{3}}$

Work Step by Step

We factor by grouping: $7x^{3}-x+1=x^{3}+3x^{2}+x$ $7x^{3}-x+1-x^{3}-3x^{2}-x=0$ $6x^{3}-3x^{2}-2x+1=0$ $3x^{2}(2x-1)-(2x-1)=0$ $(2x-1)(3x^{2}-1)=0$ $2x-1=0$ or $3x^{2}-1=0$ $x=\frac{1}{2}$ or $x=\pm\sqrt{\frac{1}{3}}$

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