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To solve \( (4 - 6i)(3 - 2i) \), we’ll use the distributive property (also known as FOIL for binomials) and then simplify:

1. **Expand the expression**:
   \[
   (4 - 6i)(3 - 2i) = 4 \cdot 3 + 4 \cdot (-2i) + (-6i) \cdot 3 + (-6i) \cdot (-2i)
   \]

2. **Calculate each term**:
   - \( 4 \cdot 3 = 12 \)
   - \( 4 \cdot (-2i) = -8i \)
   - \( (-6i) \cdot 3 = -18i \)
   - \( (-6i) \cdot (-2i) = 12i^2 \)

3. **Combine like terms and simplify**:
   - Combine the real parts and the imaginary parts separately:
     - Real part: \( 12 + 12i^2 \)
     - Imaginary part: \( -8i - 18i \)

4. **Substitute \( i^2 = -1 \)**:
   - Since \( i^2 = -1 \), \( 12i^2 = 12 \cdot (-1) = -12 \).
   - So the real part becomes \( 12 - 12 = 0 \).
   - The imaginary part is \( -8i - 18i = -26i \).

5. **Write the answer in standard form**:
   - The result is \( 0 - 26i \), or simply \( -26i \).

So, the answer in standard form is:
\[
-26i
\]

Would you like more details or have any questions?

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Here are five related questions that could help deepen understanding:

1. How would the result differ if \( i \) had been positive in both terms?
2. What happens if we multiply two complex conjugates, like \( (4 + 6i)(4 - 6i) \)?
3. How can the modulus (magnitude) of \( -26i \) be found?
4. What does it mean geometrically to multiply two complex numbers?
5. How would this result look in polar form?

**Tip:** When multiplying complex numbers, remember that \( i^2 = -1 \), which is essential for simplifying terms!

Multiply (4 - 6i)(3 - 2i). Write your answer as a complex number in standard form.

Learn how to multiply complex numbers (4 - 6i)(3 - 2i) and express the result in standard form. This solution covers key concepts in complex number operations and the imaginary unit. Suitable for Algebra 2 students in Grades 10-11.

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