Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 60



Work Step by Step

As per given question: $(x-6)(x+2)=20$ This can be written as: $x^2+2x-6x-12-20=0$ or, $x^2-4x-32=0$ Need to use factorization to solve the quadratic equation . $x^2+4x-8x-32=0$ or, $(x+4)(x-8)=0$ This implies $x+4=0$ and $x-8=0$ Hence, $x=-4,8$
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