To solve the multiplication of two complex numbers \((3 + i) \cdot (-2 - 3i)\), follow these steps:

1. **Apply the distributive property** (also known as the FOIL method for binomials):

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

2. **Multiply the terms**:

\[
3 \cdot (-2) = -6
\]
\[
3 \cdot (-3i) = -9i
\]
\[
i \cdot (-2) = -2i
\]
\[
i \cdot (-3i) = -3i^2
\]

3. **Simplify** \(i^2 = -1\):

\[
i^2 = -1, \quad \therefore -3i^2 = -3(-1) = 3
\]

4. **Combine like terms**:

\[
-6 + 3 - 9i - 2i = (-6 + 3) + (-9i - 2i)
\]
\[
= -3 - 11i
\]

So, the result is:

\[
(3 + i) \cdot (-2 - 3i) = -3 - 11i
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. What is the geometric interpretation of multiplying two complex numbers?
2. How do you find the modulus of a complex number like \(3 + i\)?
3. What is the conjugate of a complex number, and how does it affect multiplication?
4. How do you add two complex numbers, such as \(3 + i\) and \(−2 − 3i\)?
5. How does the polar form of complex numbers simplify multiplication?

**Tip:** When multiplying complex numbers, always remember that \(i^2 = -1\), which is key to simplifying terms.

Explore how to multiply two complex numbers, (3 + i) and (−2 − 3i), using the distributive property. This solution includes detailed steps, multiplication of terms, and simplifying i^2 to arrive at the final result of -3 - 11i. Ideal for students in Grades 10-12.

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