ALGEBRA 2 question!?

Question

Perform the following complex number multiplication and write the answer in standard form:

(3 - 2 i)(-2 - 4 i)

Answer 1

You need to use the "FOIL" method to solve this.
F=first (the 3 and the -2)
O=outside (the 3 and the -4i)
I=inside (-2i and -2)
L=last (-2i and -41)

F: 3 * -2 = -6
O: 3 * -4i = -12i
I: -2i * -2 = 4i
L: -2i * -4i = 8i^2

So.....
(3 - 2i)(-2 - 4i) = -6 -12i + 4i + 8i^2
= -6 - 8i + 8(-1)
= -14 - 8i

Answer 2

Using the FOIL (First, Outside, Inside, Last) you will expand the complex number.
(3-2i)(-2-4i)
First: 3(-2)=-6
Outside: 3(-4i)=-12i
Inside: -2i(-2)= 4i
Last: -2i(-4i)= 8i^2

then combine like terms

-6-8i+8i^2

recall that i^2=-1

-6-8i+8(-1)
-6-8i-8
-14-8i

your answer in standard form is

-14-8i

Answer 3

I love doing these... he he.
Use the foil method... which stands for the order that you can multiply to make sure you do them all. First Out Inner Last

(3-2i)(-2-4i)
F= -6
O= -12i
I= 4i
L= 8i^2

then add them all (i like to put the x^2 first then the x then the whole numbers)
8i^2 - 8i - 6

simplify by divided them all by 2
4i^2 - 4i - 3

Answer 4

= (3 - 2i) (- 2 - 4i)
= - 6 + 4i - 12i + 8i^2
= 8i^2 - 8i -6

Answer: 8i^2 - 8i - 6

Answer 5

(3 - 2 i)(-2 - 4 i)
<==>
-2(3 - 2i)(1 + 2i)
-2(3 + 6i - 2i - 4i^2)
we know that i^2 = -1, then
-2(3 + 4i + 4)
-2(4i + 7)

Answer 6

(3 - 2 i)(-2 - 4 i)
3(-2) + 3(-4i) + (-2i)(-2) + (-2i)(-4i)
-6 + (-12i) + 4i + 8i^2
-6 - 8i + 8(-1)
-6 - 8i - 8
-14 - 8i

Answer 7

(3 - 2 i)(-2 - 4 i)
=3(-2) + 3(-4i)-2i(-2)-2i(-4i)
=-6 -12i +4i +8i^2
=14 -8i

Answer 8

-14-8i

Answer 9

=-6 -8i+8i^2
=2-8i