# How do you simplify (4-2i)(6+6i)?

Aug 26, 2016

You have to cross-multiply the two factors (4−2i)(6+6i) as if they were factors of a quadratic.

#### Explanation:

Remember FOIL? First terms, Inside terms, Outside terms, Last terms. Do this first.

Next, combine like terms.

Lastly, recognize that ${i}^{2}$ equals $- 1$, and take that into account.

So, let's multiply (4−2i)(6+6i) :
$4 \cdot 6 = 24$
$4 \cdot 6 i = 24 i$
$- 2 i \cdot 6 = - 12 i$
$- 2 i \cdot 6 i = - 12 {i}^{2}$

Since ${i}^{2} = - 1$, we have $- 12 {i}^{2} = + 12$.

Add all the answers we got above:

$24 + 24 i - 12 i + 12$

Combine the two $i$ terms:

$24 i - 12 i = + 12 i$, and so far we have:

$24 + 12 i + 12$

Combine the constants ( 24 + 12) and we get:

$36 + 12 i$.