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Every positive integer greater than 1 is either the product of two or more integer factors greater than 1, in which case it is a composite number, or it is not, in which case it is a prime number. For example, 15 is a composite number because 15 = 3 · 5, but 7 is a prime number because it cannot be decomposed in this way.
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.
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.
Then, by strong induction, assume this is true for all numbers greater than 1 and less than n. If n is prime, there is nothing more to prove. Otherwise, there are integers a and b, where n = a b, and 1 < a ≤ b < n. By the induction hypothesis, a = p 1 p 2 ⋅⋅⋅ p j and b = q 1 q 2 ⋅⋅⋅ q k are products of primes.
A polynomial f of degree n greater than one, which is irreducible over F q, defines a field extension of degree n which is isomorphic to the field with q n elements: the elements of this extension are the polynomials of degree lower than n; addition, subtraction and multiplication by an element of F q are those of the polynomials; the product ...
At this point gcd(91645-2,112729) = 811, so 811 is a non-trivial factor of 112729. Notice that p−1 = 810 = 2 × 5 × 3 4. The number 9! is the lowest factorial which is multiple of 810, so the proper factor 811 is found in this step. The factor 139 is not found this time because p−1 = 138 = 2 × 3 × 23 which is not a divisor of 9!
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If the field F is not algebraically closed, then the minimal and characteristic polynomials need not factor according to their roots (in F) alone, in other words they may have irreducible polynomial factors of degree greater than 1. For irreducible polynomials P one has similar equivalences: P divides μ A, P divides χ A,
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