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: 44


$x=\left\{ 1,4 \right\}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $ (x+2)(x-7)=-18 ,$ is equivalent to \begin{array}{l}\require{cancel} x(x)+x(-7)+2(x)+2(-7)=-18 \\\\ x^2-7x+2x-14=-18 \\\\ x^2-7x+2x-14+18=0 \\\\ x^2-5x+4=0 .\end{array} Factoring the above equation, $ x^2-5x+4=0 ,$ results to \begin{array}{l}\require{cancel} (x-4)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (x-4)(x-1)=0 ,$ are \begin{array}{l}\require{cancel} x-4=0 \\\\ x=4 ,\\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $ x=\left\{ 1,4 \right\} .$
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