Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set - Page 473: 61


$y=0$ or $y=-1$ or $y=-2$

Work Step by Step

$ y^{3}+3y^{2}+2y=0\qquad$...factor out the common term, $y$. $y(y^{2}+3y+2)=0$ ... Searching for two factors of $ac=2$ whose sum is $b=3,$ we find$\qquad 1$ and $2.$ Rewrite the middle term and factor in pairs: $y(y^{2}+y+2y+2)=0$ $y[y(y+1)+2(y+1)]=0$ $ y(y+1)(y+2)=0\qquad$...apply the principle of zero products. $y=0$ $y+1=0$ $y=-1$ $y+2=0$ $y=-2$ Type the equation into a graphing utility and see that the graph intercepts the x-axis at $-1,-2$ and $0$.
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