can someone help me with i......(3 + 2i)(2 + 4i)?

You treat the imaginary number like any pronumeral in this case. (just remember that i^2 = -1)

Therefore:

(3+2i)(2+4i)
6 + 12i + 4i + 8i^2 (simply expand everything from the brackets)
6 + 16i +8(-1) (simplify, remembering to change i^2 to its equal value of -1)
6+ 16i - 8 (simplify further)
= 16i -2 (TA-DA!!!)

That's it really. All you need to know is that an imaginary number is the square root of a negative but since a negative number can't really be 'square-rooted' we simply assign it the value of 'i'. 'i' is the square root of -1 so therefor i^2 (i squared) is -1.

(3 + 2i)(2 + 4i)
= 3(2) + 3(4i) + 2i(2) + 2i(4i)
= 6 + 12i + 4i + 8i^2
= 6 + 16i + 8(-1)
= 6 + 16i - 8
= -2 + 16i

6 + 4i + 12i + 8i^2

6 + 16i -8

-2 + 16i

i^2 = -1

i² = -1
6+16i+8 i² = 16i-2

One thing you need to know is that i² = -1. And so...

(3 + 2i)(2 + 4i)

FOIL as usual

6 + 12i + 4i + 8i²

6 + 16i + 8(-1)

6 + 16i - 8

-2 + 16i

:)

just normal expanding

6+12i+4i+8i^2
= 6 +16i +8(-1)
= -2+16i

i^2=-1

surprised by the number of people getting it wrong

6 + 12 i + 4 i + 8 i ²

6 + 16 i - 8

16 i - 2

(3 + 2i)(2 + 4i)
=6+12i+4i+8i^2
=6+18i+8(-1) note i^2=-1
=6+18i-8
=-2+18i answer//

easy... 3+2i 2+4i =
5i 6i

altogether.... 11i ....:)

(3 + 2i)(2 + 4i)

= 6 + 4i + 12i + 8i^2
= 6 + 16i + 8i^2
= 6 + 16(-1) + 8(-1)^2
= 6 - 16 +8(1)
= -10 + 8
= -2