So asking " Factorise this quadratic equation " isn't going to work, but saying " I'm having trouble understanding how to factorise quadratic equations " is fine, and people will point you at stuff to read and non-wiki tutorials .-- Talk 20 : 18, 15 October 2006 ( UTC)
22.
So asking " Factorise this quadratic equation " isn't going to work, but saying " I'm having trouble understanding how to factorise quadratic equations " is fine, and people will point you at stuff to read and non-wiki tutorials .-- Talk 20 : 18, 15 October 2006 ( UTC)
23.
If you're asked to solve it, then you would probably factorise it first anyway, and then use that to find the solutions-( p + a ) ( p + b ) = 0 means that either p + a = 0 or p + b = 0, and then solve like any linear equation.
24.
For example to factorise a 2 + ab-ab 2-b 3, Can I treat it as a function of a-ie . f ( a ) = a 2 + ab-ab 2-b 3 and then say that since f ( b 2 ) = 0, then ( a-b 2 ) is a factor of this expression.
25.
I can see the method is closely related to Dixon's method, but as I said I fail to see why the CFRAC method is actually advantageous-does it perhaps typically find that the A n 2 ( using the terminology of the linked CFRAC explanation ) are congruent to a very small number, in which case more likely to factorise under a set of small primes?
26.
:Not much of answer, but here is something you CAN do : you can calculate the product of the first few thousand prime numbers ( say all the primes less than 10, 000 ) and store it as "'N "', and then when you have a big number "'M "'you want to factorise, you can start by finding the GCD of "'N "'and "'M " '.
27.
I've been learning ways of factorising numbers through the Congruence of Squares method, and at the moment I'm looking at the continued fraction method, whereby you use the convergents of a continued fraction expansion of Sqrt ( N ) in order to find a congruence of squares modulo N : I think http : / / www . math . dartmouth . edu / ~ carlp / PDF / implementation . pdf section 2 has one of the few good descriptions of the process I can find online.
28.
Not every number is an appropriate choice for the SNFS : you need to know in advance a polynomial " f " of appropriate degree ( the optimal degree is conjectured to be \ left ( 3 \ frac { \ log N } { \ log \ log N } \ right ) ^ { 1 / 3 }, which is 4, 5, or 6 for the sizes of N currently feasible to factorise ) with small coefficients, and a value " x " such that f ( x ) \ equiv 0 \ pmod N where N is the number to factorise.
29.
Not every number is an appropriate choice for the SNFS : you need to know in advance a polynomial " f " of appropriate degree ( the optimal degree is conjectured to be \ left ( 3 \ frac { \ log N } { \ log \ log N } \ right ) ^ { 1 / 3 }, which is 4, 5, or 6 for the sizes of N currently feasible to factorise ) with small coefficients, and a value " x " such that f ( x ) \ equiv 0 \ pmod N where N is the number to factorise.
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