To solve \((2 - i)(-3 + 6i)\), we'll use the distributive property (FOIL method) again.

\[
(2 - i)(-3 + 6i) = 2(-3) + 2(6i) - i(-3) - i(6i)
\]

Now, let's calculate each part:

1. \(2 \times -3 = -6\)  (First)
2. \(2 \times 6i = 12i\)  (Outer)
3. \(-i \times -3 = 3i\)  (Inner)
4. \(-i \times 6i = -6i^2\)  (Last)

Next, combine like terms and simplify:

\[
-6 + 12i + 3i - 6i^2
\]

Combine the real and imaginary parts:

\[
-6 + 15i - 6i^2
\]

Since \(i^2 = -1\), substitute \(-1\) for \(i^2\):

\[
-6 + 15i - 6(-1) = -6 + 15i + 6
\]

Finally, combine the real parts:

\[
-6 + 6 + 15i = 0 + 15i
\]

So, the product is:

\[
\boxed{15i}
\]

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. What is the magnitude of the complex number \(15i\)?
2. What happens when you multiply two purely imaginary numbers?
3. How can you express \(15i\) in polar form?
4. What are the properties of multiplication of complex conjugates?
5. How does the multiplication of complex numbers relate to rotation in the complex plane?

**Tip:** When multiplying complex numbers, carefully track the signs, especially when dealing with the imaginary unit \(i\) where \(i^2 = -1\).

Learn how to multiply complex numbers (2 - i)(-3 + 6i) using the distributive property (FOIL method). This problem involves complex numbers and the imaginary unit, with a step-by-step solution provided. Suitable for students in Grades 10-12.

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