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To solve the given expression \( 3i(2 + 4i) - 5i(6 + 2i) \), we can expand each term separately and then combine them. Let's go through the steps:

### Step 1: Expand \( 3i(2 + 4i) \)
\[
3i(2 + 4i) = 3i \cdot 2 + 3i \cdot 4i
\]
\[
= 6i + 12i^2
\]
Since \( i^2 = -1 \), we substitute \( i^2 \) with \(-1\):
\[
= 6i + 12(-1) = 6i - 12
\]

### Step 2: Expand \( -5i(6 + 2i) \)
\[
-5i(6 + 2i) = -5i \cdot 6 - 5i \cdot 2i
\]
\[
= -30i - 10i^2
\]
Again, substitute \( i^2 = -1 \):
\[
= -30i - 10(-1) = -30i + 10
\]

### Step 3: Combine the results
Now we add the results from Step 1 and Step 2:
\[
(6i - 12) + (-30i + 10)
\]
Combine like terms:
\[
= -24i - 2
\]

### Final Answer
The expression simplifies to \(-2 - 24i\), which matches option **A**.

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Would you like further details or have any questions?

### Related Questions:
1. How do you distribute complex numbers in expressions?
2. What is the significance of \(i^2 = -1\) in complex number multiplication?
3. How can you simplify expressions with multiple complex terms?
4. What other properties of imaginary numbers are useful in algebra?
5. How can you verify if a solution to a complex equation is correct?

### Tip:
Remember, when multiplying complex numbers, always simplify \( i^2 \) to \(-1\) to make combining terms easier.

Which expression is equivalent to 3i(2 + 4i) - 5i(6 + 2i)?

Explore how to simplify the complex expression 3i(2 + 4i) - 5i(6 + 2i) by expanding and combining terms. Learn about imaginary units and algebraic operations with complex numbers. Suitable for high school students in Grades 10-12.

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