Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 6 - Factoring, Solving Equations, and Problem Solving - Chapter 6 Test - Page 275: 16



Work Step by Step

First, we factor out $x$ from the expression as it is common to all the terms: $x^{3}+16x^{2}+48x=0$ $x(x^{2}+16x+48)=0$ Using the rules of factoring trinomials to factor the polynomial, we obtain: $x(x^{2}+16x+48)=0$ $x(x^{2}+4x+12x+48)=0$ $x[x(x+4)+12(x+4)]=0$ $x(x+4)(x+12)=0$ Now, we equate the factors to zero: $x(x+4)(x+12)=0$ $x=0$ or $(x+4)=0$ or $(x+12)=0$ $x=0$ or $x=-4$ or $x=-12$ Therefore, the solution set is {$-12,-4,0$}.
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