How do you factor completely $2y^3+8y^2-64y$ ?

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Answer 1 · 2015-06-13T14:45:26

Factors are 2, y, y+8, y-4

Please, have a look how the given figure can be broken down:
$2y^3+8y^2−64y = 2y(y^2+4y-32) = 2y(y^2+8y-4y-32) = 2y{y(y+8)-4(y+8) = 2y(y+8)(y-4)$

How do you factor completely 2y^3+8y^2-64y2y^3+8y^2-64y?