Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 356: 42


$\left\{ -\dfrac{8}{5},0,\dfrac{8}{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 25x^3=64x ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 25x^3-64x=0 .\end{array} The $GCF$ of the constants of the terms $\{ 25,-64 \}$ is $ 1 $ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ x^3,x \}$ is $ x .$ Hence, the entire expression has $GCF= x .$ Factoring the $GCF= x ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(25x^2-64)=0 .\end{array} The expressions $ 25x^2 $ and $ 64 $ are both perfect squares and are separated by a minus sign. Hence, $ 25x^2-64 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(5x+8)(5x-8)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 5x+8=0 \\\\\text{OR}\\\\ 5x-8=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 5x+8=0 \\\\ 5x=-8 \\\\ x=-\dfrac{8}{5} \\\\\text{OR}\\\\ 5x=8 \\\\ x=\dfrac{8}{5} .\end{array} Hence, the solutions are $ \left\{ -\dfrac{8}{5},0,\dfrac{8}{5} \right\} .$
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