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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).
In mathematics, Ruffini's rule is a method for computation of the Euclidean division of a polynomial by a binomial of the form x – r. It was described by Paolo Ruffini in 1809. [ 1 ] The rule is a special case of synthetic division in which the divisor is a linear factor.
Start with division by 2: the number is even, and n = 2 · 693. Continue with 693, and 2 as a first divisor candidate. 693 is odd (2 is not a divisor), but is a multiple of 3: one has 693 = 3 · 231 and n = 2 · 3 · 231. Continue with 231, and 3 as a first divisor candidate. 231 is also a multiple of 3: one has 231 = 3 · 77, and thus n = 2 ...
Polynomial factoring algorithms use basic polynomial operations such as products, divisions, gcd, powers of one polynomial modulo another, etc. A multiplication of two polynomials of degree at most n can be done in O(n 2) operations in F q using "classical" arithmetic, or in O(nlog(n) log(log(n)) ) operations in F q using "fast" arithmetic.
An optimal strategy for choosing these polynomials is not known; one simple method is to pick a degree d for a polynomial, consider the expansion of n in base m (allowing digits between −m and m) for a number of different m of order n 1/d, and pick f(x) as the polynomial with the smallest coefficients and g(x) as x − m.
The formula for the difference of two squares can be used for factoring polynomials that contain the square of a first quantity minus the square of a second quantity. For example, the polynomial x 4 − 1 {\displaystyle x^{4}-1} can be factored as follows:
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We have s(P) = s(1,1) = 4, so the coordinates of 2P = (x ′, y ′) are x ′ = s 2 – 2x = 14 and y ′ = s(x – x ′) – y = 4(1 – 14) – 1 = –53, all numbers understood (mod n). Just to check that this 2P is indeed on the curve: (–53) 2 = 2809 = 14 3 + 5·14 – 5. Then we compute 3(2P). We have s(2P) = s(14,-53) = –593/106 ...
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