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

Question

3500 views around the world

You can reuse this answer
Creative Commons License

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#.