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: 62


$y=-3$ or $y=1$ or $y=0$

Work Step by Step

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