How to solve (-4+4i) (6+3i)?

(-4 + 4i)(6 + 3i)
= -4(6) - 4(3i) + 4i(6) + 4i(3i)
= -24 - 12i + 24i + 12i^2
= -24 + 12i + 12(-1)
= -24 + 12i - 12
= -24 - 12 + 12i
= -36 + 12i

Remembering that i^2 = -1, (i-4)(i+2) + (3i-6) = (-4 + i)(2 + i) + (-6 + i)..............(writing the expression in standard x + iy form) = -8 - 4i + 2i + i^2 - 6 + i = -8 - i -1 = -9 - i

- 24 - 12 i + 24 i + 12 i ²

- 24 + 12 i - 12

- 36 + 12 i

12 ( - 3 + i )

(-4+4i)(6+3i)
multiply
-24-12i+24i+12i^2
simplify
12i^2+12i-24
note that i=squareroot of -1 so i^2=-1
12(-1)+12i-24
simplify
12i-36

-36+12i

by multiplying, we get, -24+24i-12i+12i^2

this implies, -24+12i-12 (since i^2 = -1)

therefore, 12i-36

or 36-12i is the answer, where i = (-1)^1/2

epic fail, it factoring but with i. lets see
12(-1+i)(2+i) = 12(-2-i+2i+i^2) = 12(-3+i) = -36+12i

note that i^2= -1

if you write this expression in simple form, you'll have:
12(-1+i)(2+i) = 12(-2-i+2i+i^2) = 12(-3+i) = -36+12i