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

Jan 19, 2018

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

#### 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.

$\left(- 3 i\right) \left(2 i\right) \left(8 + 2 i\right)$

We can start with $\left(- 3 i\right) \left(2 i\right)$, which becomes $- 6 {i}^{2}$:

$= \left(- 6 {i}^{2}\right) \left(8 + 2 i\right)$

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

$= 8 \left(- 6 {i}^{2}\right) + 2 i \left(- 6 {i}^{2}\right)$

$= - 48 {i}^{2} + - 12 {i}^{3}$

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

$= - 48 {i}^{2} + - 12 {i}^{2} i$

We know that by definition, ${i}^{2} = - 1$:

$= - 48 \left(- 1\right) + - 12 \left(- 1\right) i$

$= 48 + 12 i$

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

$\left(- 3 i\right) \left(2 i\right) \left(8 + 2 i\right) = 48 + 12 i$.