Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises - Page 361: 50


$x=\left\{ -2,-\dfrac{6}{5},3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x+2)(5x^2-9x-18)=0 ,$ express it first in factored form. Then equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 5(-18)=-90 $ and the value of $b$ is $ -9 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 6,-15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} (x+2)(5x^2+6x-15x-18)=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+2)[(5x^2+6x)-(15x+18)]=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} (x+2)[x(5x+6)-3(5x+6)]=0 .\end{array} Factoring the $GCF= (5x+6) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x+2)[(5x+6)(x-3)]=0 \\\\ (x+2)(5x+6)(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x+2=0 \\\\\text{OR}\\\\ 5x+6=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+2=0 \\\\ x=-2 \\\\\text{OR}\\\\ 5x+6=0 \\\\ 5x=-6 \\\\ x=-\dfrac{6}{5} \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, $ x=\left\{ -2,-\dfrac{6}{5},3 \right\} .$
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