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The image presents a mathematical expression involving products and factors of odd integers. Let's break it down:

1. The first line shows the product \( x \cdot y = (2k + 1)(2k' + 1) \), which is the product of two odd numbers, where \( k \) and \( k' \) are integers.
2. Expanding this product, we apply distributive property:
   \[
   (2k + 1)(2k' + 1) = 4kk' + 2k + 2k' + 1
   \]
   This is the second line in the image.
3. The third line factorizes this expression:
   \[
   4kk' + 2k + 2k' + 1 = 2(2kk' + k + k') + 1
   \]
   This highlights that the expression is of the form \( 2m + 1 \), meaning it is odd, as shown by the label "impar" (odd).

The image divides the expression into a sum of an even part \( 2(2kk' + k + k') \) and an odd constant \( 1 \).

Would you like more detailed steps or further explanation on any part? Here are a few related questions for practice:

1. How can you prove that the product of two odd numbers is always odd?
2. What happens when you multiply two even numbers? Does the result have a similar structure?
3. Can you generalize this approach for more complex odd integer products?
4. How does this relate to number theory and modular arithmetic?
5. How does factorizing expressions like these help in understanding properties of integers?

**Tip**: Always check for patterns in the parity (even or odd nature) of numbers when simplifying algebraic expressions involving integers. This can help simplify your work significantly!

Given the product of two odd numbers (2k + 1)(2k' + 1), simplify the expression and determine its parity.

Explore how to simplify the product of two odd numbers (2k + 1)(2k' + 1) using algebraic factorization. Understand how to determine its parity and the relevance of parity rules in algebra. This problem involves key algebraic concepts such as factorization and parity analysis, suitable for students in Grades 9-12.

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