<s>[INST] Can you give me a generalized form to find all the roots of a polynomial to the n-th degree using the newton method? [/INST] The generalized form of the Newton Method to find the roots of a polynomial to the n-th degree is given by the following equation: 

Xk+1 = Xk - (f(Xk)/f'(Xk))

where Xk is an initial guess and f(Xk) is the value of the polynomial evaluated at Xk, and f'(Xk) is the derivative of the polynomial evaluated at Xk.</s>\ <s>[INST] Can you give me examples? [/INST] ### Example 1: Find a root of the quadratic function f(x) = x^2 - 4

Define the function and its derivative:
f(x) = x^2 - 4
f'(x) = 2x

Choose an initial guess, say X0 = 1.

Apply the Newton-Raphson formula:
X1 = X0 - (f(X0) / f'(X0)) = 1 - ((1^2 - 4) / (2 * 1)) = 1 + 1.5 = 2.5

Iterate the process until the desired precision is reached:
X2 = X1 - (f(X1) / f'(X1)) = 2.5 - ((2.5^2 - 4) / (2 * 2.5)) ≈ 2.05

The method converges quickly, and after a few iterations, you'll find that the root is approximately 2, which is correct since the exact root is 2.

### Example 2: Find a root of the cubic function f(x) = x^3 - 6x^2 + 11x - 6

Define the function and its derivative:
f(x) = x^3 - 6x^2 + 11x - 6
f'(x) = 3x^2 - 12x + 11

Choose an initial guess, say X0 = 1.

Apply the Newton-Raphson formula:
X1 = X0 - (f(X0) / f'(X0)) = 1 - ((1 - 6 + 11 - 6) / (3 - 12 + 11)) = 1 - 0 = 1

Iterate the process until the desired precision is reached:
X2 = X1 - (f(X1) / f'(X1)) = 1 - ((1 - 6 + 11 - 6) / (3 - 12 + 11)) = 1 - 0 = 1

In this case, the method converged to the root in just one iteration, which is exactly 1. Note that the Newton-Raphson method may not always find all roots, and its success depends on the initial guess and the function's properties.</s>