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: 44


$\left\{ -\dfrac{1}{2},-7,7 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 2x^3+x^2-98x-49=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+x^2)-(98x+49)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(2x+1)-49(2x+1)=0 .\end{array} Factoring the $GCF= (2x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+1)(x^2-49)=0 .\end{array} The expressions $ x^2 $ and $ 49 $ are both perfect squares and are separated by a minus sign. Hence, $ x^2-49 ,$ 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+1)(x+7)(x-7)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 2x+1=0 \\\\\text{OR}\\\\ x+7=0 \\\\\text{OR}\\\\ x-7=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+1=0 \\\\ 2x=-1 \\\\ x=-\dfrac{1}{2} \\\\\text{OR}\\\\ x+7=0 \\\\ x=-7 \\\\\text{OR}\\\\ x-7=0 \\\\ x=7 .\end{array} Hence, the solutions are $ \left\{ -\dfrac{1}{2},-7,7 \right\} .$
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