Method for Solving Polynomial Equations

The purpose of my paper is to bring a method for solving polynomial equations using basic algebra and series and also using combinatorics. A series which converges to the solutions of polynomial equations. The contribution of this method is that it leads directly to precise results to find the roots of a polynomial equation of any degree starting from second degree to infinity and also for the solving of radicals since radicals are a particular type of polynomial equations for example to find the square root of 2 sends to solve the equation x2=2. A general formula for the series which converges to the solutions of polynomial equations. For complex solutions we write for a polynomial P(x), P(a+bi)=P(a-bi)=0 and to solve this separately for imaginary part and real part of the solution sends to solve for regular polynomial equations at one variable so we can use the method which is developed to find the solutions. Citation: Julien Y (2018) Method for Solving Polynomial Equations. J Appl Computat Math 7: 409. doi: 10.4172/2168-9679.1000409


Particular cases:
* When A(n0-1)=0 Before being able to solve the equation, we first need to transform the equation into a classical n0 order equation when at least the coefficient of x n0-1 and x 0 will be different from 0. This can be done by setting X=x+a , then we can solve the new equation in x.
Example: Let's get a look at the equation X 3 =3X+4 Substitution of X by x+a where " a" can be any real leads to: so developing the cubic form leads to: x 3 +a 3 +3 a 2 x+3a x 2 =3x+3a+4 this leads to: We need to transform the equation for i by setting X=x+a and then we need to recalculate the next terms of the series E(k).

Complex Solutions
For a polynomial P(X)=0,P(a+bi)=P(abi)=0 so we need to solve separately for the real parts and imaginary parts so it sends to solve a regular polynomial equation at one variable to find a and b
So let's formulate a general result for the number of iterations needed to calculate the limit L.
Let it be a no th polynomial equation: The number of iterations needed is determined by the minimum common factor Ko.
Then for m digits the number of iterations that will always give the m digits is In the decimal system K+10 iterations will be the maximum number of iterations needed that the m digits are also common to the subsequent terms of the serie G(n).   For calculating Ko, a rule is that only the A(i)≠0 are taken into account for the product. In any case the minimum condition is A(n0-1)×A(0)≠0.
and also real integers.
Then we can express the product for Ko are taken into account for the product. In any case the minimum condition is Now the question which can be raised up is how to know the number of digits which a rational solution may have.
Determination of the maximum number of digits for a rational solution in the decimal base Let it be a n0 th polynomial equation If X has m digits in the decimal base so the number of digits of the left side can be expressed Am+B and in the right side can be expressed Cm+D,So since the left side is equal to the right side so Am+B=Cm+D (A-C)*m=D-B so m=(D-B)/(A-C) The biggest value for m can be obtained for D maximum, B minimum, A minimum, C maximum C maximum=n0-1A minimum=n0B minimum=-n0+1 When M(k) is the number of digits for A(k): example if A(n0-1)=13 so M(n0-1)=2.
What Henrik Niels Abel has demonstrated is that a general formula for the solution of polynomial equations can not be obtained in terms of radicals for degree n ≥ 5 .
However the equation X 2 =2 is in itself a mathematical problem to solve because saying that the solution of this equation is 2 X = does not give the arithmetic expression of the solution In order to solve the equation X 2 =2 we need first to transform this equation into a classical second order equation. This can be done by transformation with X=x+a (x+a) 2 =2 this leads to: x 2 +a 2 +2ax=2⇔ x 2 =-2ax+2-a 2.
One example that will lead to the solution can be a=1 this leads to x 2 =-2x+2-1=-2x+1 so we need to solve the equation.
x 2 =-2x+1 then X=x+1 corresponds to the solution of X 2 =2 or in other terms we can express the square root of 2 : Let's solve the equation Let's express the recurrent relationship for E(n) and G(n) Before we can use those formulas we need first to calculate the first terms for E(n) and Applying the formula for G(n) leads to: and F is a function defined as this: We can rewrite the formula for E(k): The other solution is x 2 =2.414213562-2=0.414213562 So those are the two square roots of 2 So my conclusion is that since we can find the radicals in terms of simple operations like additions multiplications subtractions divisions so we should be able to find general formulas which can be expressed in terms of simple operations to express the solutions of any nth polynomial equation. All what is needed for this is to find explicit formulas for the series E(n) and G(n).
So we need to identify the different pair of prime factors contained in the product between the five products.
[  In the decimal system K+10=5+10=15 interactions is the maximum number of the interactions for which m=1 digits after the decimal point is also common to the subsequent terms of the series which means that L(with m=1 digit)=G(15)=G (16) Infact checking the result with a calculator will show that gives 9 digits after the decimal point of the solution x.
So X (with 9 digits after the decimal point)=[int(G(15)*10 9 )]/10 9 Where int is the integers function Now we need to calculate E(15) and E(14) using the expression of E(k) developed for the quantic formula we need to remember that only positive powers are allowed in the expression.
We calculate E(14) So k=14 ,a=3,b=4,c=2,d=1,e=1 Let's express the root of a quantic equation by using the expression of E k developed for the quinitic formula (n 0 =5) The first thing we need to determine is the minimum number of interaction needed .to calculate the solution of the equation. This number is called K 1 =K0+n0 following the derivatives given to determine Kc.
We needed E(15) and E (14) For k0+k1+k2+k3>7 the powers of a=3 in E(15) will be negative and since the condition in the E(k)formula is K(no-1)≥ 0 , which means that the powers of a is to be positive powers .So we need to step the summation k0+k1+k2+k3 at the sum equal to 7 to calculate E(15).