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If the polynomial has rational roots, for example x 2 − 4x + 4 = (x − 2) 2, or x 2 − 3x + 2 = (x − 2)(x − 1), then the Galois group is trivial; that is, it contains only the identity permutation. In this example, if A = 2 and B = 1 then A − B = 1 is no longer true when A and B are swapped.
It follows from the present theorem and the fundamental theorem of algebra that if the degree of a real polynomial is odd, it must have at least one real root. [2] This can be proved as follows. Since non-real complex roots come in conjugate pairs, there are an even number of them; But a polynomial of odd degree has an odd number of roots;
Even for the first root that involves at most two square roots, the expression of the solutions in terms of radicals is usually highly complicated. However, when no square root is needed, the form of the first solution may be rather simple, as for the equation x 5 − 5 x 4 + 30 x 3 − 50 x 2 + 55 x − 21 = 0 , for which the only real solution is
An illustration of Newton's method. In numerical analysis, the Newton–Raphson method, also known simply as Newton's method, named after Isaac Newton and Joseph Raphson, is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function.
Alexandrov's uniqueness theorem (discrete geometry) Alperin–Brauer–Gorenstein theorem (finite groups) Alspach's theorem (graph theory) Amitsur–Levitzki theorem (linear algebra) Analyst's traveling salesman theorem (discrete mathematics) Analytic Fredholm theorem (functional analysis) Anderson's theorem (real analysis)
In modular arithmetic, a number g is a primitive root modulo n if every number a coprime to n is congruent to a power of g modulo n. That is, g is a primitive root modulo n if for every integer a coprime to n, there is some integer k for which g k ≡ a (mod n). Such a value k is called the index or discrete logarithm of a to the base g modulo n.
Vieta's formulas are frequently used with polynomials with coefficients in any integral domain R.Then, the quotients / belong to the field of fractions of R (and possibly are in R itself if happens to be invertible in R) and the roots are taken in an algebraically closed extension.
If the rational root test finds no rational solutions, then the only way to express the solutions algebraically uses cube roots. But if the test finds a rational solution r, then factoring out (x – r) leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.