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

To find the negative root of the polynomial \( x^3 - 2x^2 - 5x + 6 = 0 \) using the Newton-Raphson method, follow these steps:

1. **Identify the function and its derivative:**
   \[
   f(x) = x^3 - 2x^2 - 5x + 6
   \]
   \[
   f'(x) = 3x^2 - 4x - 5
   \]

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

3. **Initial guess:**
   To find the negative root, choose an initial guess \( x_0 \) that is negative. Let's start with \( x_0 = -1 \).

4. **Iterate using the Newton-Raphson formula:**
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]

   - **Iteration 1:**
     \[
     x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -1 - \frac{(-1)^3 - 2(-1)^2 - 5(-1) + 6}{3(-1)^2 - 4(-1) - 5}
     \]
     \[
     f(-1) = -1 - 2 + 5 + 6 = 8
     \]
     \[
     f'(-1) = 3(1) + 4 - 5 = 2
     \]
     \[
     x_1 = -1 - \frac{8}{2} = -1 - 4 = -5
     \]

   - **Iteration 2:**
     \[
     x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -5 - \frac{(-5)^3 - 2(-5)^2 - 5(-5) + 6}{3(-5)^2 - 4(-5) - 5}
     \]
     \[
     f(-5) = -125 - 50 + 25 + 6 = -144
     \]
     \[
     f'(-5) = 3(25) + 20 - 5 = 90
     \]
     \[
     x_2 = -5 - \frac{-144}{90} = -5 + 1.6 = -3.4
     \]

   - **Iteration 3:**
     \[
     x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = -3.4 - \frac{(-3.4)^3 - 2(-3.4)^2 - 5(-3.4) + 6}{3(-3.4)^2 - 4(-3.4) - 5}
     \]
     \[
     f(-3.4) = -39.304
     \]
     \[
     f'(-3.4) = 28.54
     \]
     \[
     x_3 = -3.4 - \frac{-39.304}{28.54} \approx -3.4 + 1.377 = -2.023
     \]

   - **Iteration 4:**
     \[
     x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = -2.023 - \frac{(-2.023)^3 - 2(-2.023)^2 - 5(-2.023) + 6}{3(-2.023)^2 - 4(-2.023) - 5}
     \]
     \[
     f(-2.023) = -6.403
     \]
     \[
     f'(-2.023) = 13.558
     \]
     \[
     x_4 = -2.023 - \frac{-6.403}{13.558} \approx -2.023 + 0.472 = -1.551
     \]

   - **Iteration 5:**
     \[
     x_5 = x_4 - \frac{f(x_4)}{f'(x_4)} = -1.551 - \frac{(-1.551)^3 - 2(-1.551)^2 - 5(-1.551) + 6}{3(-1.551)^2 - 4(-1.551) - 5}
     \]
     \[
     f(-1.551) = -0.921
     \]
     \[
     f'(-1.551) = 6.488
     \]
     \[
     x_5 = -1.551 - \frac{-0.921}{6.488} \approx -1.551 + 0.142 = -1.409
     \]

   - **Iteration 6:**
     \[
     x_6 = x_5 - \frac{f(x_5)}{f'(x_5)} = -1.409 - \frac{(-1.409)^3 - 2(-1.409)^2 - 5(-1.409) + 6}{3(-1.409)^2 - 4(-1.409) - 5}
     \]
     \[
     f(-1.409) = -0.057
     \]
     \[
     f'(-1.409) = 5.405
     \]
     \[
     x_6 = -1.409 - \frac{-0.057}{5.405} \approx -1.409 + 0.011 = -1.398
     \]

5. **Conclusion:**
   After several iterations, the value converges to approximately \( x \approx -1.398 \), which is the negative root of the equation \( x^3 - 2x^2 - 5x + 6 = 0 \).

Would you like more details or have any questions about this process?

### Additional Questions:
1. What is the general approach to the Newton-Raphson method?
2. How do you determine a good initial guess for the Newton-Raphson method?
3. Why is it important to find the derivative of the function when using the Newton-Raphson method?
4. How does the convergence of the Newton-Raphson method depend on the initial guess?
5. What are the potential pitfalls of the Newton-Raphson method?
6. Can the Newton-Raphson method be used for functions other than polynomials?
7. How can you check the accuracy of the root found using the Newton-Raphson method?
8. Are there any alternatives to the Newton-Raphson method for finding roots of equations?

### Tip:
Always verify the initial guess and perform a few iterations manually to ensure that the Newton-Raphson method is converging towards the correct root.

Learn how to find the negative root of the polynomial x³ - 2x² - 5x + 6 using the Newton-Raphson method. Follow a step-by-step guide with detailed iterations and calculations. Suitable for advanced high school students and beyond.

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