To solve \((2 + 3i) \times (-1 + 4i)\) using the distributive property, we expand as follows:

\[
(2 + 3i)(-1 + 4i) = 2 \cdot (-1) + 2 \cdot (4i) + 3i \cdot (-1) + 3i \cdot (4i)
\]

Now calculate each term:

1. \( 2 \cdot (-1) = -2 \)
2. \( 2 \cdot 4i = 8i \)
3. \( 3i \cdot (-1) = -3i \)
4. \( 3i \cdot 4i = 12i^2 \)

Since \( i^2 = -1 \), we replace \( 12i^2 \) with \( 12(-1) = -12 \).

Now combine all the terms:

\[
-2 + 8i - 3i - 12 = (-2 - 12) + (8i - 3i)
\]

Simplify:

\[
-14 + 5i
\]

Thus, the result of \((2 + 3i) \times (-1 + 4i)\) is:

\[
-14 + 5i
\]

Would you like any further details on this process?

Here are 5 related questions to expand this topic:

1. How do you multiply two complex numbers in polar form?
2. What is the geometric interpretation of multiplying two complex numbers?
3. How do you divide complex numbers?
4. What is the modulus of the product of two complex numbers?
5. How do you convert a complex number from rectangular to polar form?

**Tip:** Always remember that \( i^2 = -1 \) when working with imaginary numbers.

Learn how to multiply complex numbers using the distributive property with a detailed step-by-step solution for (2 + 3i) × (-1 + 4i). This problem covers key concepts in complex numbers and algebra, including the distributive property and handling imaginary units (i). Suitable for students in Grades 10-12.

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