Answer
$$3y(y+2)(y-8) $$
Work Step by Step
We take out $3y$ from the given polynomial
$$3y^3 - 18y^2 - 48y=3y(y^2 -6y -16)$$
To factor the remaining polynomial, we have to find integers whose product is $-16$ and whose sum is $-6$. That is, we have $2$ and $-8$, where $2-8=-6$ and $2\times (-8)=-16$. Thus $$ 3y^3 - 18y^2 - 48y=3y(y+2)(y-8) .$$
