Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.2 - Factoring the Difference of Two Squares - Problem Set 6.2: 68

Answer

The solution set is $\left\{-\frac{1}{4}, 0, \frac{1}{4}\right\}$.

Work Step by Step

Add $-4x$ to both sides to find: $64x^3-4x=0$ Factor out $4x$ to find: $4x(16x^2-1)=0 \\=4x[(4x)^2-1^{2}]=0$ RECALL: A difference of two squares can be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares to find: $4x(4x-1)(4x+1)=0$ Equate each factor to 0 then solve each equation to find: $4x=0 \text{ or } 4x-1=0 \text{ or } 4x+1=0 \\x=0 \text{ or } 4x=1 \text{ or } 4x=-1 \\x=0 \text{ or } x=\frac{1}{4} \text{ or } x=-\frac{1}{4}$ The solution set is $\left\{-\frac{1}{4}, 0, \frac{1}{4}\right\}$.
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