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We are asked to solve the cubic equation:

\[
x^3 + 2x^2 - 5x - 6 = 0
\]

with the given information that \(x = 2\) is a root.

### Steps:

1. **Verify the root:**  
   Substitute \(x = 2\) into the equation:

   \[
   2^3 + 2(2^2) - 5(2) - 6 = 8 + 8 - 10 - 6 = 0
   \]

   Since this holds true, \(x = 2\) is indeed a root.

2. **Factor the polynomial using synthetic division:**  
   Since \(x = 2\) is a root, we can divide the cubic polynomial by \(x - 2\).

   Perform synthetic division on \(x^3 + 2x^2 - 5x - 6\) by \(x - 2\):

   ```
   2 | 1   2   -5   -6
       |     2    8    6
       ----------------
         1   4    3    0
   ```

   The quotient is \(x^2 + 4x + 3\), and the remainder is 0.

3. **Factor the quadratic:**  
   Now we solve the quadratic \(x^2 + 4x + 3\) by factoring:

   \[
   x^2 + 4x + 3 = (x + 1)(x + 3)
   \]

4. **Find all the roots:**  
   The full factorization of the original polynomial is:

   \[
   (x - 2)(x + 1)(x + 3) = 0
   \]

   Therefore, the solutions are:

   \[
   x = 2, \quad x = -1, \quad x = -3
   \]

### Final Answer:
The solutions to the equation \(x^3 + 2x^2 - 5x - 6 = 0\) are:

\[
x = 2, \quad x = -1, \quad x = -3
\]

Would you like further details or have any questions?

---

Here are five related questions to explore further:

1. How does synthetic division work, and why is it useful in solving cubic equations?
2. What other methods can be used to solve cubic equations if a root is not given?
3. Can you always factor a cubic polynomial into linear factors if all roots are real?
4. How can the quadratic formula be applied if the quadratic part isn't factorable?
5. What are some real-world applications of solving cubic equations?

**Tip:** Always verify your roots by substituting them back into the original equation!

Solve the equation x^3 + 2x^2 - 5x - 6 = 0 given that 2 is a zero of f(x).

Learn how to solve the cubic equation x^3 + 2x^2 - 5x - 6 = 0 with a given root x = 2. This problem involves synthetic division, factoring, and finding all real roots of the polynomial. Suitable for Grades 10-12.

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