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: 51



Work Step by Step

Given: $x(x+1)^3-42(x+1)^2=0$ Need to use factorization to solve the quadratic equation. This can be factorized as follows: $(x+1)^2(x^2+x-42)=0$ Now $(x+1)^2=0$ and $(x^2+x-42)=0$ This implies $x=-1$ and $(x^2+x-42)=0$ or, $x^2-6x+7x-42=0$ or, $(x-6)(x+7)=0$ This implies $x=6,-7$ Hence, $x=-1,6,7$
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