College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.6 - Factoring - P.6 Exercises: 95

Answer

$(y-3)(y-2)(y+2)$

Work Step by Step

$y^{3}-3y^{2}-4y+12 =(y^{3}-3y^{2})+(-4y+12)=y^{2}(y-3)+(-4)(y-3)=(y-3)(y^{2}-4)$ Last we use $(a^2-b^2)=(a+b)(a-b)$: $=(y-3)(y-2)(y+2)$
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