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The polynomial x 2 + cx + d, where a + b = c and ab = d, can be factorized into (x + a)(x + b). In mathematics, factorization (or factorisation, see English spelling differences) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.
The theory of finite fields, whose origins can be traced back to the works of Gauss and Galois, has played a part in various branches of mathematics.Due to the applicability of the concept in other topics of mathematics and sciences like computer science there has been a resurgence of interest in finite fields and this is partly due to important applications in coding theory and cryptography.
The smaller piece, at the bottom, has width a-b and height b. Now the smaller piece can be detached, rotated, and placed to the right of the larger piece. In this new arrangement, shown in the last diagram below, the two pieces together form a rectangle, whose width is a + b {\displaystyle a+b} and whose height is a − b {\displaystyle a-b} .
Modern algorithms and computers can quickly factor univariate polynomials of degree more than 1000 having coefficients with thousands of digits. [3] For this purpose, even for factoring over the rational numbers and number fields, a fundamental step is a factorization of a polynomial over a finite field.
Construct an ambiguous form (a, b, c) that is an element f ∈ G Δ of order dividing 2 to obtain a coprime factorization of the largest odd divisor of Δ in which Δ = −4ac or Δ = a(a − 4c) or Δ = (b − 2a)(b + 2a). If the ambiguous form provides a factorization of n then stop, otherwise find another ambiguous form until the ...
In mathematics, an irreducible polynomial is, roughly speaking, a polynomial that cannot be factored into the product of two non-constant polynomials.The property of irreducibility depends on the nature of the coefficients that are accepted for the possible factors, that is, the ring to which the coefficients of the polynomial and its possible factors are supposed to belong.
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If a and b have different parity, let p be any factor of a 2 + b 2 such that p 2 < a 2 + b 2. Then c = a 2 + b 2 − p 2 / 2p and d = a 2 + b 2 + p 2 / 2p . Note that p = d − c. A similar method exists [5] for generating all Pythagorean quadruples for which a and b are both even. Let l = a / 2 and m = b / 2 and ...