To factor the quadratic expression \( 2n^2 + 3n - 9 \) completely, we can use the method of "factor by grouping."

### Steps:

1. **Multiply the leading coefficient and the constant term**:
   \[
   2 \times (-9) = -18
   \]
2. **Find two numbers that multiply to give \(-18\) and add to give the middle coefficient \(3\)**:
   \[
   6 \times (-3) = -18 \quad \text{and} \quad 6 + (-3) = 3
   \]
3. **Rewrite the middle term \(3n\) as the sum of \(6n\) and \(-3n\)**:
   \[
   2n^2 + 6n - 3n - 9
   \]
4. **Group the terms in pairs**:
   \[
   (2n^2 + 6n) - (3n + 9)
   \]
5. **Factor out the greatest common factor from each group**:
   \[
   2n(n + 3) - 3(n + 3)
   \]
6. **Factor out the common binomial factor \((n + 3)\)**:
   \[
   (2n - 3)(n + 3)
   \]

Thus, the factored form of \( 2n^2 + 3n - 9 \) is:
\[
(2n - 3)(n + 3)
\]

Would you like further explanation on any of these steps?

### Related Questions:
1. How do you factor quadratics when the leading coefficient is 1?
2. What is the difference between factoring by grouping and using the quadratic formula?
3. Can all quadratic expressions be factored by grouping?
4. How can you check if a factorization is correct?
5. What is the role of the discriminant in factoring quadratics?

**Tip**: Always check your factorization by expanding the factors to make sure you get back to the original expression!

Learn how to factor the quadratic expression 2n^2 + 3n - 9 completely using the method of factor by grouping. This problem covers key algebra concepts such as factoring quadratics and identifying common factors. Suitable for students in Grades 8-10.

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