https://file.aiquickdraw.com/mathbot/file/1729791646238zlf5k8xs.jpg

Let's work through the factoring problems from the image.

### Question 14: Factor \( m^2 + 5mn + 6n^2 \)

This is a quadratic trinomial in terms of \( m \). To factor it, we need two numbers that multiply to give \( 6n^2 \) and add up to \( 5n \).

- Factors of \( 6n^2 \) are \( (2n) \) and \( (3n) \).
- These add to \( 5n \).

So, the factorization is:
\[
m^2 + 5mn + 6n^2 = (m + 2n)(m + 3n)
\]

### Question 15: Factor \( m^2 - 11mn + 28n^2 \)

Again, this is a quadratic trinomial in terms of \( m \). We need two numbers that multiply to \( 28n^2 \) and add up to \( -11n \).

- Factors of \( 28n^2 \) are \( (-7n) \) and \( (-4n) \).
- These add to \( -11n \).

Thus, the factorization is:
\[
m^2 - 11mn + 28n^2 = (m - 7n)(m - 4n)
\]

Let me know if you'd like more detailed explanations!

Here are 5 related questions to expand on these problems:
1. How do you factor trinomials where the leading coefficient isn't 1?
2. What are the steps to identify the greatest common factor (GCF) before factoring?
3. How do you solve factored quadratic equations?
4. What is the difference between factoring by grouping and simple trinomial factoring?
5. Can every trinomial be factored, or are some prime?

**Tip:** Always check if you can factor out a common factor from all terms before applying other factoring techniques!

Factor m^2 + 5mn + 6n^2 and Factor m^2 - 11mn + 28n^2

Learn how to factor the quadratic expressions m^2 + 5mn + 6n^2 and m^2 - 11mn + 28n^2 with detailed step-by-step solutions. This problem covers key algebraic concepts suitable for students in Grades 8-10.

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