The expression you provided, \( 734_{8} \times 4_{8} \), suggests multiplication in base 8 (octal system).

### Step 1: Convert octal numbers to decimal (base 10)
1. Convert \( 734_8 \) to base 10:
   \[
   734_8 = 7 \times 8^2 + 3 \times 8^1 + 4 \times 8^0 = 7 \times 64 + 3 \times 8 + 4 = 448 + 24 + 4 = 476_{10}
   \]
2. Convert \( 4_8 \) to base 10:
   \[
   4_8 = 4_{10}
   \]

### Step 2: Multiply in base 10
Now, multiply the two decimal numbers:
\[
476_{10} \times 4_{10} = 1904_{10}
\]

### Step 3: Convert the result back to base 8
To convert \( 1904_{10} \) to base 8:
- \( 1904 \div 8 = 238 \) remainder \( 0 \)
- \( 238 \div 8 = 29 \) remainder \( 6 \)
- \( 29 \div 8 = 3 \) remainder \( 5 \)
- \( 3 \div 8 = 0 \) remainder \( 3 \)

Thus, \( 1904_{10} = 3560_8 \).

### Final Answer:
\[
734_8 \times 4_8 = 3560_8
\]

Would you like further details on any step? Here are related questions you might find interesting:

1. How do you convert other base numbers to base 10?
2. What are the properties of octal numbers?
3. How can we do multiplication directly in base 8 without converting to decimal?
4. How does division work in the octal system?
5. What is the significance of number systems in computing?

**Tip:** When converting between number bases, always start by breaking down powers of the base to simplify the conversion process.

Learn how to multiply octal numbers by converting to decimal, performing the multiplication, and converting the result back to octal. This problem demonstrates base conversion and multiplication concepts, suitable for Grades 9-12.

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