Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 352: 49


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

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $ (x+3)(3x+5)=7 ,$ is equivalent to \begin{array}{l}\require{cancel} x(3x)+x(5)+3(3x)+3(5)=7 \\\\ 3x^2+5x+9x+15=7 \\\\ 3x^2+5x+9x+15-7=0 \\\\ 3x^2+14x+8=0 .\end{array} Factoring the above equation, $ 3x^2+14x+8=0 ,$ results to \begin{array}{l}\require{cancel} (x+4)(3x+2)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (x+4)(3x+2)=0 ,$ are \begin{array}{l}\require{cancel} x+4=0 \\\\ x=-4 ,\\\\\text{OR}\\\\ 3x+2=0 \\\\ 3x=-2 \\\\ x=-\dfrac{2}{3} .\end{array} Hence, $ x=\left\{ -4,-\dfrac{2}{3} \right\} .$
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