How do you simplify #(8y^3-16y^2)/(2y^3-2y^2-4y)#?

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Answer 1 · 2015-06-16T23:51:05

You can start by eliminating extra #y# factors.

#(8y^3 - 16y^2)/(2y^3 - 2y^2 - 4y)#

#= (cancel(y)(8y^2 - 16y))/(cancel(y)(2y^2 - 2y - 4))#

Now how about those #2# factors?
#= (cancel(2)(4y^2 - 8y))/(cancel(2)(y^2 - y - 2))#

And any canceling? Yup.

#= (4ycancel((y - 2)))/(cancel((y - 2))(y + 1))#

#= (4y)/(y + 1)#