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:

$3y(y-4)(y+4)$

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