# How do you FOIL (4z-u)(3z+2u)?

Mar 24, 2018

The FOIL'ed version of the factored expression is $12 {z}^{2} + 5 z u - 2 {u}^{2}$

#### Explanation:

As you may know, FOIL stands for First Outer Inner Last. Using this Mnemonic, we'll work in order.

First:
$4 z \times 3 z = 4 \times 3 \times z \times z = 12 {z}^{2}$

Outer:
$4 z \times 2 u = 4 \times 2 \times z \times u = 8 z u$

Inner:
$- u \times 3 z = - 1 \times 3 \times u \times z = - 3 u z = - 3 z u$

Last:
$- u \times 2 u = - 1 \times 2 \times u \times u = - 2 {u}^{2}$

Now, let's put them back together in the expression:

$12 {z}^{2} + 8 z u - 3 z u - 2 {u}^{2}$

Finally we combine like terms to get our finished expression:

$12 {z}^{2} + \left(8 - 3\right) z u - 2 {u}^{2} = \textcolor{red}{12 {z}^{2} + 5 z u - 2 {u}^{2}}$

Mar 24, 2018

$12 {z}^{2} + 5 z u - 2 {u}^{2}$

#### Explanation:

The FOIL method:

1. (4z−u)(3z+2u)
2. $\left(4 z \cdot 3 z\right) + \left(4 z \cdot 2 u\right) + \left(- u \cdot 3 z\right) + \left(- u \cdot 2 u\right)$
3. $12 {z}^{2} + 8 z u - 3 z u - 2 {u}^{2}$
4. $12 {z}^{2} + 5 z u - 2 {u}^{2}$

Hope this helps!

Mar 24, 2018

$2$${x}^{2}$+$5$$u$$z$$-$$2$${u}^{2}$

#### Explanation:

$F$= First
$O$= Outside
$I$= Inside
$L$= Last

$\left(4 z - u\right)$ $\left(3 z + 2 u\right)$

Since $4 z \mathmr{and} 3 z$ are first you multiply them by each other...

$4 \cdot 3$ equals $12$
$z \cdot z$ equals ${z}^{2}$

$4 z \mathmr{and} 2 u$ are at the end so you would multiply them together
and get $8 u z$ it can also be $8 z u$ but later on you learn that the letter that comes first in the alphabet is usually at the beginning.

The inside would be
$- u \mathmr{and} 3 z$
In which $- u \cdot 3 z$ would equal $- 3 u z$

Finally, you solve the last two...
$- u \cdot 2 u$
Which equals $- 2 {u}^{2}$

You're equation should now look like this;
$12 {z}^{2} + 8 u z - 3 u z - 2 {u}^{2}$

Subtract $8 u z \mathmr{and} 3 u z$ - They have common variables
You should get $5 u z$

Your final equation should look like this;
$12 {z}^{2} + 5 u z - 2 {u}^{2}$