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Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.
The question was: How to solve a polynomial of degree n. Knowing to where to find the solution is an answer to the question cited. In general, there are no exact solutions for solving polynomials in terms of radicals, that is in terms of square roots, cube roots , etc., for polynomials of degree five or greater, and the solutions are ...
3. As mentioned above, no general formula to find all the roots of any 5th degree equation exists, but various special solution techniques do exist. My own favourite: - By inspection, see if the polynomial has any simple real solutions such as x = 0 or x = 1 or -1 or 2 or -2. If so, divide the poly by (x-a), where a is the found root, and then ...
5. Since complex number field C is algebraically closed, every polynomials with complex coefficients have linear polynomial decomposition. In this case, it's z3 − 3z2 + 6z − 4 = (z − 1)(z − 1 + √3i)(z − 1 − √3i). So you can see the solution of the equation easily from this representation. One way to find out such decomposition ...
Because your equation can be rearranged to a polynomial exponential equation over the algebraic numbers, it cannot have solutions that are elementary numbers. Because your exponential polynomial equation has a polynomial term of more than one non-constant monomials of different degree, Lambert W cannot applied here.
In any case, if you start with a system of polynomial equations and compute a Groebner basis for the ideal they generate, you get a "maximally triangular" system of equations which is equivalent to the original one---that is why Groebner bases generalize Gaussian elimination. Let me do a simple example using Macaulay 2. Consider the ring ...
4. One important consequence of being able to explicitly solve polynomial equations is that it permits great simplifications by linearizing what would otherwise be much more complicated nonlinear phenomena. The ability to factor polynomials completely into linear factors over C C enables widespread linearization simplifications of diverse problems.
In case this was your actual question I will expand the comments a little: Given any polynomial P P in x x with real coefficients, a general quaternion solution will have the form q = a +b1i +b2j +b3k q = a + b 1 i + b 2 j + b 3 k with a,b1,b2,b3 ∈ R a, b 1, b 2, b 3 ∈ R. If q q not real let.
Solving polynomials in one variable over finite fields is substantially easier than solving polynomials in general. To find out if f(x) = 0 f (x) = 0 has any roots over Fq F q you just need to compute gcd(f(x),xq − x) gcd (f (x), x q − x) using the Euclidean algorithm, since the roots of xq − x x q − x are precisely the elements of Fq F q.
You can also solve a quadratic equation by the technique of completing the square. So if you need to solve the equation x2 + 8x = 5 in the ring Z17, then you can add 16 to both sides to get x2 + 8x + 16 = 5 + 16 = 21 ≡ 4 (mod 17). This you can write in the form (x + 4)2 = 4 =22. So x + 2 must be an element with square equal to 4.