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Dividing f(x) by p(x) gives the other factor () = +, so that () = (). Now one can test recursively to find factors of p ( x ) and q ( x ), in this case using the rational root test. It turns out they are both irreducible, so the irreducible factorization of f ( x ) is: [ 5 ]
Algorithm: SFF (Square-Free Factorization) Input: A monic polynomial f in F q [x] where q = p m Output: Square-free factorization of f R ← 1 # Make w be the product (without multiplicity) of all factors of f that have # multiplicity not divisible by p c ← gcd(f, f′) w ← f/c # Step 1: Identify all factors in w i ← 1 while w ≠ 1 do y ...
For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorization is not usually considered meaningful within number systems possessing division , such as the real or complex numbers , since any x {\displaystyle x} can be trivially written as ( x y ) × ( 1 / y ) {\displaystyle ...
In general, not every GCD in the above product will be a non-trivial factor of (), but some are, providing the factors we seek. Berlekamp's algorithm finds polynomials g ( x ) {\displaystyle g(x)} suitable for use with the above result by computing a basis for the Berlekamp subalgebra.
In algebraic number theory, the Dedekind–Kummer theorem describes how a prime ideal in a Dedekind domain factors over the domain's integral closure. [1] It is named after Richard Dedekind who developed the theorem based on the work of Ernst Kummer .
Decomposition: = where C is an m-by-r full column rank matrix and F is an r-by-n full row rank matrix Comment: The rank factorization can be used to compute the Moore–Penrose pseudoinverse of A , [ 2 ] which one can apply to obtain all solutions of the linear system A x = b {\displaystyle A\mathbf {x} =\mathbf {b} } .
Hensel's original lemma concerns the relation between polynomial factorization over the integers and over the integers modulo a prime number p and its powers. It can be straightforwardly extended to the case where the integers are replaced by any commutative ring, and p is replaced by any maximal ideal (indeed, the maximal ideals of have the form , where p is a prime number).
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...