# How do you FOIL (3x^2-6)(x^3-9)?

May 31, 2015

$\left(3 {x}^{2} - 6\right) \left({x}^{3} - 9\right) = 3 {x}^{5} - 27 {x}^{2} - 6 {x}^{3} + 54$

Problem: FOIL $\left(3 {x}^{2} - 6\right) \left({x}^{3} - 9\right)$ .

FOIL stands for:

First: Multiply the first term in each binomial together.
Outer/Outside: Multiply the outer/outside term in each binomial together.
Inner/Inside: Multiply the inner/inside term in each binomial together.
Last: Multiply the last term in each binomial together.

$\left(3 {x}^{2} - 6\right) \left({x}^{3} - 9\right)$ =

$\left(3 {x}^{2} \cdot {x}^{3}\right) + \left(3 {x}^{2} \cdot - 9\right) + \left(- 6 \cdot {x}^{3}\right) + \left(- 6 \cdot \left(- 9\right)\right)$ =

$3 {x}^{5} - 27 {x}^{2} - 6 {x}^{3} + 54$