Factor completely with solution  1) 2y²+2.     2) m3+343.           3) 3m³+24

Let's factor each expression completely:

### 1) \( 2y^2 + 2 \)

First, factor out the common factor:
\[
2(y^2 + 1)
\]
This is fully factored since \(y^2 + 1\) cannot be factored further using real numbers.

### Final Answer: 
\[
2(y^2 + 1)
\]

---

### 2) \( m^3 + 343 \)

This is a sum of cubes. We use the formula for factoring a sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
Here, \(m^3 + 343\) can be written as:
\[
m^3 + 7^3
\]
Using the sum of cubes formula:
\[
m^3 + 343 = (m + 7)(m^2 - 7m + 49)
\]

### Final Answer: 
\[
(m + 7)(m^2 - 7m + 49)
\]

---

### 3) \( 3m^3 + 24 \)

First, factor out the common factor of \(3\):
\[
3(m^3 + 8)
\]
Now, factor \(m^3 + 8\), which is a sum of cubes:
\[
m^3 + 2^3 = (m + 2)(m^2 - 2m + 4)
\]
Substitute this back into the expression:
\[
3(m + 2)(m^2 - 2m + 4)
\]

### Final Answer: 
\[
3(m + 2)(m^2 - 2m + 4)
\]

---

### Do you have any questions or would you like more details?

### Related Questions:
1. How do you factor a difference of cubes?
2. Can you explain how to recognize a sum of cubes pattern?
3. What happens when factoring higher-degree polynomials?
4. How do you factor trinomials like \(x^2 + bx + c\)?
5. What are some techniques for factoring by grouping?

### Tip: 
For cube factorizations, remember the formula for both the sum and difference of cubes, as these often appear in polynomial problems.

Factor completely with solution 1) 2y²+2. 2) m³+343. 3) 3m³+24.

Learn how to factor the polynomials 2y²+2, m³+343, and 3m³+24. This solution covers key algebraic concepts like factoring using the greatest common factor and the sum of cubes theorem. Perfect for Grades 9-10 students studying polynomial factorization.

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