To find a root of the equation $x^3 - 2x^2 - 5x + 6 = 0$ using the Newton-Raphson method, follow these steps:

1. Define the function and its derivative: Let $f(x) = x^3 - 2x^2 - 5x + 6$. The derivative of $f(x)$ is $f'(x) = 3x^2 - 4x - 5$.

2. Newton-Raphson iteration formula: The Newton-Raphson iteration formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

3. Choose an initial guess: We need an initial guess, $x_0$. Let's start with $x_0 = 2$.

4. Iterate until convergence: We perform iterations until the values converge to a desired level of accuracy.

First iteration:

$x_0 = 2$ $f(2) = 2^3 - 2 \cdot 2^2 - 5 \cdot 2 + 6 = 8 - 8 - 10 + 6 = -4$ $f'(2) = 3 \cdot 2^2 - 4 \cdot 2 - 5 = 12 - 8 - 5 = -1$ $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-4}{-1} = 2 - 4 = -2$

Second iteration:

$x_1 = -2$ $f(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 = -8 - 8 + 10 + 6 = 0$ $f'(-2) = 3(-2)^2 - 4(-2) - 5 = 12 + 8 - 5 = 15$ $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -2 - \frac{0}{15} = -2$

We see that $x_1 = -2$ yields $f(-2) = 0$, indicating that $x = -2$ is a root of the equation.

Verification:

Substitute $x = -2$ back into the original equation: $(-2)^3 - 2(-2)^2 - 5(-2) + 6 = -8 - 8 + 10 + 6 = 0$ Thus, $x = -2$ is indeed a root.

Conclusion:

The root found using the Newton-Raphson method is $x = -2$.

Would you like more details on any of these steps, or do you have any questions?

Here are 8 related questions that expand on this information:

1. Can the Newton-Raphson method find all roots of a polynomial equation?
2. What are the conditions for the Newton-Raphson method to converge?
3. How can we choose a good initial guess for the Newton-Raphson method?
4. Are there any cases where the Newton-Raphson method might fail?
5. Can we use the Newton-Raphson method for non-polynomial equations?
6. How does the rate of convergence for the Newton-Raphson method compare to other methods?
7. What are the advantages and disadvantages of the Newton-Raphson method compared to the fixed point iteration method?
8. Can the Newton-Raphson method be used for complex roots?

Tip: The Newton-Raphson method converges quadratically near a root, making it very efficient. However, a poor initial guess can lead to divergence or convergence to a different root.