## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$$3y(y+2)(y-8)$$
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 $$3y^3 - 18y^2 - 48y=3y(y+2)(y-8) .$$