Find the possible root of x³ - 2x²-5x+6=0 using the newton raphson method 

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.

Learn how to find a root of the polynomial equation x³ - 2x² - 5x + 6 using the Newton-Raphson method. Detailed step-by-step solution with initial guess, iterations, and verification included. Suitable for advanced high school students and beyond.

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