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  2. Factorization of polynomials over finite fields - Wikipedia

    en.wikipedia.org/wiki/Factorization_of...

    The polynomial P = x 4 + 1 is irreducible over Q but not over any finite field. On any field extension of F 2, P = (x + 1) 4. On every other finite field, at least one of −1, 2 and −2 is a square, because the product of two non-squares is a square and so we have; If =, then = (+) ().

  3. Factorization of polynomials - Wikipedia

    en.wikipedia.org/wiki/Factorization_of_polynomials

    If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as 1×2, 2×1, (−1)×(−2), or (−2)×(−1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values p(0) = 1, 2, −1, or −2. and likewise for p(-1

  4. Factor theorem - Wikipedia

    en.wikipedia.org/wiki/Factor_theorem

    In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f ( x ) {\displaystyle f(x)} is a polynomial, then x − a {\displaystyle x-a} is a factor of f ( x ) {\displaystyle f(x)} if and only if f ( a ) = 0 {\displaystyle f(a)=0} (that is, a {\displaystyle a} is a root of the polynomial).

  5. Factorization - Wikipedia

    en.wikipedia.org/wiki/Factorization

    The next odd divisor to be tested is 7. One has 77 = 7 · 11, and thus n = 2 · 3 2 · 7 · 11. This shows that 7 is prime (easy to test directly). Continue with 11, and 7 as a first divisor candidate. As 7 2 > 11, one has finished. Thus 11 is prime, and the prime factorization is; 1386 = 2 · 3 2 · 7 · 11.

  6. Table of Gaussian integer factorizations - Wikipedia

    en.wikipedia.org/wiki/Table_of_Gaussian_Integer...

    The entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right complex half plane with absolute value of the real part larger than or equal to the absolute value of the imaginary part.

  7. Fermat's factorization method - Wikipedia

    en.wikipedia.org/wiki/Fermat's_factorization_method

    Squares are always congruent to 0, 1, 4, 5, 9, 16 modulo 20. The values repeat with each increase of a by 10. In this example, N is 17 mod 20, so subtracting 17 mod 20 (or adding 3), produces 3, 4, 7, 8, 12, and 19 modulo 20 for these values. It is apparent that only the 4 from this list can be a square.

  8. Polynomial - Wikipedia

    en.wikipedia.org/wiki/Polynomial

    For example, antiderivatives of x 2 + 1 have the form ⁠ 1 / 3x 3 + x + c. For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p , or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ...

  9. Complex conjugate root theorem - Wikipedia

    en.wikipedia.org/wiki/Complex_conjugate_root_theorem

    The polynomial x 2 + 1 = 0 has roots ± i. Any real square matrix of odd degree has at least one real eigenvalue. For example, if the matrix is orthogonal, then 1 or −1 is an eigenvalue. The polynomial + has roots , +,, and thus can be factored as