Simplify (2+3i)⋅(1−4i)(2 + 3i) \cdot (1 - 4i)(2+3i)⋅(1−4i).

We are tasked with simplifying the expression \((2 + 3i) \cdot (1 - 4i)\). Let's go step by step.

### Step 1: Use the distributive property (FOIL method)
We distribute the terms:
\[
(2 + 3i)(1 - 4i) = 2(1) + 2(-4i) + 3i(1) + 3i(-4i)
\]
Now simplify each term:
- \(2 \cdot 1 = 2\)
- \(2 \cdot (-4i) = -8i\)
- \(3i \cdot 1 = 3i\)
- \(3i \cdot (-4i) = -12i^2\)

### Step 2: Simplify using \(i^2 = -1\)
Recall that \(i^2 = -1\), so:
\[
-12i^2 = -12(-1) = 12
\]

### Step 3: Combine all terms
Now substitute back into the expression:
\[
2 - 8i + 3i + 12
\]
Combine the real terms and the imaginary terms:
\[
(2 + 12) + (-8i + 3i) = 14 - 5i
\]

### Final Answer:
\[
(2 + 3i) \cdot (1 - 4i) = 14 - 5i
\]

Would you like more details on this process or clarification on any step?

Here are 5 related questions you could explore:
1. What happens when you multiply two complex numbers geometrically?
2. How do you find the conjugate of a complex number?
3. Can you simplify \((3 + 2i)(2 + 5i)\)?
4. What is the modulus of the complex number \(14 - 5i\)?
5. How do you divide two complex numbers?

**Tip:** When multiplying complex numbers, always remember that \(i^2 = -1\), which simplifies the terms with imaginary units.

This guide walks through the step-by-step solution to simplify the multiplication of complex numbers (2+3i)(1−4i). Learn key concepts, including distributing terms and using the formula i^2 = -1, suitable for students in Grades 9-12.

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