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

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Answer 1 · 2016-08-26T19:29:36

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

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 $(4−2i)(6+6i)$ :
$4 * 6 = 24$
$4 * 6i = 24i$
$- 2i * 6 = - 12i$
$- 2i * 6i = - 12i^2$

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

Add all the answers we got above:

$24 + 24i - 12i +12$

Combine the two $i$ terms:

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

$24 + 12i + 12$

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

$36 + 12i$ .

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