Question

How do you solve `(-3i)(2i)(8+2i)`?

Answer 1

Multiply them together as if `i` is a variable, and then evaluate for `i` using `i^2 = -1` to get `(-3i)(2i)(8 + 2i) = 48 + 12i`.

Explanation:

Well, we can simply multiply them as we would with the reals, first treating `i` as a variable, and then evaluate for its value later on.

`(-3i)(2i)(8 + 2i)`

We can start with `(-3i)(2i)`, which becomes `-6i^2`:

`= (-6i^2)(8 + 2i)`

Then use the distributive property to "multiply over" the first term to the second term:

`= 8(-6i^2) + 2i(-6i^2)`

`= -48i^2 + -12i^3`

Now we can start evaluating. Let's first "split" `i^3` into `i^2` and `i`:

`= -48i^2 + -12 i^2 i`

We know that by definition, `i^2 = -1`:

`= -48(-1) + -12(-1)i`

`= 48 + 12i`

Ah, that's as far as we can go! Therefore,

`(-3i)(2i)(8 + 2i) = 48 + 12i`.