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: 43


$\left\{ -\dfrac{5}{2},-1,1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 2x^3+5x^2-2x-5=0 ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (2x^3+5x^2)-(2x+5)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(2x+5)-(2x+5)=0 .\end{array} Factoring the $GCF= (2x+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+5)(x^2-1)=0 .\end{array} The expressions $ x^2 $ and $ 1 $ are both perfect squares and are separated by a minus sign. Hence, $ x^2-1 ,$ 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} (2x+5)(x+1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 2x+5=0 \\\\\text{OR}\\\\ x+1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} \\\\\text{OR}\\\\ x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, the solutions are $ \left\{ -\dfrac{5}{2},-1,1 \right\} .$
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