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Asked by Sam

Factor completely
Y^3+y^2-4y-4
12 years ago

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Answered by Reiny
use grouping

= y^2(y+1)- 4(y+1)
= (y+1)(y^2 - 4) ---- now a difference of squares
= (y+1)(y+2)(y-2)
12 years ago
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Factor completely
Y^3+y^2-4y-4

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