Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 48


$x=\left\{ -\dfrac{7}{2},0,4 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2x^3-x^2-28x=0 ,$ express it in the factored form. Then equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation. $\bf{\text{Solution Details:}}$ Factoring the $GCF=x$, the equation above is equivalent to \begin{array}{l}\require{cancel} x(2x^2-x-28)=0 .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $ 2(-28)=-56 $ and the value of $b$ is $ -1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-56 \}, \{ 2,-28 \}, \{ 4,-14 \}, \{ 7,-8\}, \\ \{ -1,56 \}, \{ -2,28 \}, \{ -4,14 \}, \{ -7,8\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 7,-8 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x(2x^2+7x-8x-28)=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} x[(2x^2+7x)-(8x+28)]=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x[x(2x+7)-4(2x+7)]=0 .\end{array} Factoring the $GCF= (2x+7) $ of the entire expression above results to \begin{array}{l}\require{cancel} x[(2x+7)(x-4)]=0 \\\\ x(2x+7)(x-4)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 2x+7=0 \\\\\text{OR}\\\\ x-4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 2x+7=0 \\\\ 2x=-7 \\\\ x=-\dfrac{7}{2} \\\\\text{OR}\\\\ x-4=0 \\\\ x=4 .\end{array} Hence, $ x=\left\{ -\dfrac{7}{2},0,4 \right\} .$
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