approximate the solution $x=2$ using newton's method for $P(x)=-1536+6272x-11328x^2+11872x^3-7952x^4+3528x^5-1036x^6+194x^7-21x^8+x^9$ I need help with this excercise.
I know that $x=2$ is a solution for
$$P(x)=-1536+6272x-11328x^2+11872x^3-7952x^4+3528x^5-1036x^6+194x^7-21x^8+x^9$$
I want aproximate the solution $x=2$ using newton method, but, I dont know what happen.
for any initial value using mathlab or python it jumps me to solutions 3 or 4 (solutions too),
example for $x=2,000000004$

Is it possible to solve this problem?
Im sorry for my English.

Plot: (zoom)

 A: What you need to learn is how to "peel off" or factor out roots you don't want.
Suppose you apply Newton's Method seeking roots of $f(x)=0$ and you find $x=\xi_1$ as a root that you want to factor out. Therefore you now want to apply the method to $f(x)/(x-\xi_1)=0$ with the idea that you knock out the $\xi_1$ root and force convergence elsewhere.
Apply the recursion to the modified function:
$x_{n+1}=x_n-\dfrac{f(x_n)/(x-\xi_1)}{(d/dx)(f(x_n)/(x-\xi_1))@x=x_n}$
When you apply the quotient rule to the derivative in the denominator you find that a factor of $1/(x-\xi_1)$ cancels, giving a simpler recursive form:
$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)-f(x_n)/(x_n-\xi_1)}$
Note that the remived root in this firm appears only as a modifying term in the denominator; the numerator remains the original function.
That is for factoring out one root. If you factor out more roots, that corresponds to more subtractions of $f(x_n)/(x-\xi_k)$ from the denominator, so your recursion with any number of removed roots takes the form
$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)-f(x_n)\Sigma[1/(x_n-\xi_k)]}$
where the summation is over all the roots you remove.
In successive applications with removed roots you should tailor your initial guesses so that you are on the "right side" of the roots you have factored out. If you remove a root $x=3$ and you are aiming for a smaller root, then your next starting point after removing the $x=3$ root would be a smaller value than $3$.
