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

May 18, 2016

$3 y \left(y - 4\right) \left(y + 4\right)$

#### Explanation:

Notice that 3 is a factor of 3 and 48
Also y is a factor of ${y}^{2}$ and $y$

Factoring out $3 y$ giving

$3 y \left({y}^{2} - 16\right)$

But 16 is ${4}^{2}$

$3 y \left({y}^{2} - {4}^{2}\right)$

Compare this to the example: ${a}^{2} + {b}^{2} = \left(a - b\right) \left(a + b\right)$

Using the same principle we have:

$3 y \left(y - 4\right) \left(y + 4\right)$