## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises: 50

#### Answer

$x=\left\{ -\dfrac{15}{8},-1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $4(2x+3)^2-(2x+3)-3=0 ,$ 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 $z=2x+3.$ Then the equation above is equivalent to \begin{array}{l}\require{cancel} 4z^2-z-3=0 .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $4(-3)=-12$ and the value of $b$ is $-1 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\}, \\ \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 3,-4 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4z^2+3z-4z-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} (4z^2+3z)-(4z+3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} z(4z+3)-(4z+3)=0 .\end{array} Factoring the $GCF= (2z+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (4z+3)(z-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then the solutions are \begin{array}{l}\require{cancel} 4z+3=0 \\\\\text{OR}\\\\ z-1=0 .\end{array} Since $z=2x+3,$ by back-substitution, the solutions are \begin{array}{l}\require{cancel} 4(2x+3)+3=0 \\\\\text{OR}\\\\ (2x+3)-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4(2x+3)+3=0 \\\\ 8x+12+3=0 \\\\ 8x+15=0 \\\\ 8x=-15 \\\\ x=-\dfrac{15}{8} \\\\\text{OR}\\\\ (2x+3)-1=0 \\\\ 2x+3-1=0 \\\\ 2x+2=0 \\\\ 2x=-2 \\\\ x=-\dfrac{2}{2} \\\\ x=-1 .\end{array} Hence, the solutions are $x=\left\{ -\dfrac{15}{8},-1 \right\} .$

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