How do you factor completely #3y^3 - 48y #?

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Answer 1 · 2016-05-18T09:08:06

#3y(y-4)(y+4)#

Notice that 3 is a factor of 3 and 48
Also y is a factor of #y^2# and #y#

Factoring out #3y# giving

#3y(y^2-16)#

But 16 is #4^2#

#3y(y^2-4^2)#

Compare this to the example: #a^2+b^2=(a-b)(a+b)#

Using the same principle we have:

#3y(y-4)(y+4)#