Question 522289: Factor completely:
40y^4 - 5y
I just know how to simplify.. but I do not know where to go from there...
5y(8y^3 - 1) Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 5y(8y^3 - 1)
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(8y^3 - 1) is a difference of 2 cubes, (2y)^3 - 1^3
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The difference of 2 cubes a^3 - b^3 = (a-b)*(a^2 + ab + b^2)
