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: 51


$x=\left\{ -\dfrac{2}{3},\dfrac{4}{15} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 5(3x-1)^2+3=-16(3x-1) ,$ use substitution and then the method on factoring trinomials. Then equate each factor to zero (Zero Product Property) and solve for the value of the variable. Finally use back-substitution and then solve for the value of the original variable. $\bf{\text{Solution Details:}}$ Let $z=3x-1.$ Then the equation above is equivalent to \begin{array}{l}\require{cancel} 5z^2+3=-16z \\\\ 5z^2+16z+3=0 .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $ 5(3)=15 $ and the value of $b$ is $ 16 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,15\}, \{3,5\}, \\ \{-1,-15\}, \{-3,-5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5z^2+z+15z+3=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} (5z^2+z)+(15z+3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} z(5z+1)+3(5z+1)=0 .\end{array} Factoring the $GCF= (5z+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5z+1)(z+3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} 5z+1=0 \\\\\text{OR}\\\\ z+3=0 .\end{array} Since $z=3x-1,$ by back-substitution, the solutions are \begin{array}{l}\require{cancel} 5(3x-1)+1=0 \\\\\text{OR}\\\\ (3x-1)+3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5(3x-1)+1=0 \\\\= 15x-5+1=0 \\\\= 15x-4=0 \\\\= 15x=4 \\\\= x=\dfrac{4}{15} \\\\\text{OR}\\\\ (3x-1)+3=0 \\\\ 3x-1+3=0 \\\\ 3x+2=0 \\\\ 3x=-2 \\\\ x=-\dfrac{2}{3} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{2}{3},\dfrac{4}{15} \right\} .$
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