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 351: 7



Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the given equation, $ (x+1)(x-8)=0 ,$ is \begin{array}{l} x+1=0 \\\\ x=-1 ,\\\\\text{OR}\\\\ x-8=0 \\\\ x=8 .\end{array} Hence, $ x=\{-1,8\} .$
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