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In elementary algebra, factoring a polynomial reduces the problem of finding its roots to finding the roots of the factors. Polynomials with coefficients in the integers or in a field possess the unique factorization property , a version of the fundamental theorem of arithmetic with prime numbers replaced by irreducible polynomials .
For example, if n = 171 × p × q where p < q are very large primes, trial division will quickly produce the factors 3 and 19 but will take p divisions to find the next factor. As a contrasting example, if n is the product of the primes 13729, 1372933, and 18848997161, where 13729 × 1372933 = 18848997157, Fermat's factorization method will ...
Fermat's factorization method, named after Pierre de Fermat, is based on the representation of an odd integer as the difference of two squares: =. That difference is algebraically factorable as (+) (); if neither factor equals one, it is a proper factorization of N.
If two or more factors of a polynomial are identical, then the polynomial is a multiple of the square of this factor. The multiple factor is also a factor of the polynomial's derivative (with respect to any of the variables, if several). For univariate polynomials, multiple factors are equivalent to multiple roots (over a suitable extension field).
Goldbach’s Conjecture. One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes ...
In mathematics and computer algebra the factorization of a polynomial consists of decomposing it into a product of irreducible factors.This decomposition is theoretically possible and is unique for polynomials with coefficients in any field, but rather strong restrictions on the field of the coefficients are needed to allow the computation of the factorization by means of an algorithm.
1.1 Example. 2 Proofs. ... In algebra, the factor theorem connects polynomial ... Two problems where the factor theorem is commonly applied are those of factoring a ...
The question of when this happens is rather subtle: for example, for the localization of k[x, y, z]/(x 2 + y 3 + z 5) at the prime ideal (x, y, z), both the local ring and its completion are UFDs, but in the apparently similar example of the localization of k[x, y, z]/(x 2 + y 3 + z 7) at the prime ideal (x, y, z) the local ring is a UFD but ...
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