HOW TO FIND THE REAL ROOT OF POLYNOMIAL EQUATION USING NEWTON RAPHSON METHOD?

X^11+3X^10+X^9+X-15=0.

Guest Apr 13, 2017

#1**0 **

The newton-Raphson method is meant to give you a close answer, not the exact answer. It is usually used to solve other equations, like sin(x)=2x, not polynomials, because if im not mistaken there is a way to solve polynomials.

but ill try to solve it using that method anyways:

well, the first thing you need to do, is find a value that gets "close enough" to the answer. Well, that term is not very specific, And i dont have any other way to explain it, but we need to get a pretty close answer, or it wont work.

Ill explain why:

There is a function, and we want to find its roots. imagine the function has "ups" and "downs", hills and ravines. When we use the method, We get an endless series of answers, getting closer and closer to the root of the function. But we need to make sure we got inside the ravine, or one of the members of the series might "get out" and start getting further and further, meaning we wont get the accurate solution we need :(.

So, Lets find a close value: first, lets find the derivative of the function (We will use it later):

f'(x)=(x^{11}+3x^{10}+x^{9}+x-15)'=11x^{10}+30x^{9}+9x^{8}+1.

1.11 is a pretty close value.

I spoke earlier about an endless series. What you are supposed to do is pick a high integer for n (You can use any computer program for that) and calculate the n^{th} term of the series where:

x_{0}=1.11=the number we chose.

x_{n+1}=x_{n}-f(x_{n})/f'(x_{n})

that means x_{1}=x_{0}-f(x_{0})/f'(x_{0})=1.11-f(1.11)/f'(1.11)

Heres what wikipedia has to say about that subject:

https://en.wikipedia.org/wiki/Newton%27s_method

But You dont HAVE to use that method. There is a nice method that can help you find the rational solutions of the equation, meaning it can help you factorize the equations (i already posted it):

**STEP ONE:**

we have the polynomial a_{0}+a_{1}*x+a_{2}*x^{2}+a_{3}*x^{3}+........+a_{n}*x^{n}=0. We need to multiply the polynomial by a number in a way that all coefficients will be whole numbers. Sometimes we cant do it. sometimes we can.

**STEP TWO:**

find all of the whole divisors of a_{n} and a_{0} (including the negative divisors). lets call the set of the divisors of a_{0} A and the set of the divisors of an B. there are k divisors in A and m divisors in B.

Every rational root of the polynomial=A_{i}/B_{j}, A_{i} is a member of A and B_{j} is a member of B, meaning every rational root can be expressed using one divisor from A divided by one divisor from B).

**STEP THREE**

Check for every two divisors (one from A and one from B) if A_{i}/B_{j} is a root. after a finite amount of time (unless your calculator is broken) you'll have every rational root of the polynomial.

Ehrlich
Apr 13, 2017