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!