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July 31

Prime Factorisation

While finding out all of the prime factors of a number by the prime factorisation method, do we take the smallest possible prime number or the largest possible prime number, while dividing each time? 117.194.228.194 (talk) 17:17, 31 July 2008 (UTC)[reply]

Doesn't matter, either will work. I usually start with small ones since it's easier to see if they divide the number. --Tango (talk) 17:23, 31 July 2008 (UTC)[reply]

Right. The small ones are easier and hopefully when you divide by all the small factors your number is small enough that it is easier to find the larger prime factors, if there are any. Oded (talk) 17:46, 31 July 2008 (UTC)[reply]

Minor quibble, but prime factorisation is just another name for finding out all of the prime factors of a number. The method I think you're talking about is trial division. As the article mentions, there are faster ways of factoring large numbers than trying every possible factor in whatever order. -- BenRG (talk) 18:07, 31 July 2008 (UTC)[reply]

And just to be explicit, the reason why it doesn't matter which prime factor you choose is the fundamental theorem of arithmetic. —Keenan Pepper 03:32, 1 August 2008 (UTC)[reply]

117.194.228.194: consider number=a*b*c*d*e*f. As long as you divide abcdef until one factor remains, you will always get the complete list of numbers. For speed, just divide by whatever factor you think of first. --Bowlhover (talk) 11:43, 3 August 2008 (UTC)[reply]

Algebra refresher

In the equation, (-2X^3)(-X^2)(2X), do you add the exponents together or multiply them? This sample test I have doesn't explain how I get to the answer of 4X^6. Thanks, Dismas|^(talk) 23:44, 31 July 2008 (UTC)[reply]

You add. Xⁿ is just a shorthand for X multiplied by itself n times, so (X²)(X³) (say) is (XX)(XXX)=XXXXX=X⁵. It's clear that multiplying can't be right, since in that case multiplying by X¹=X wouldn't change the exponent. Algebraist 23:50, 31 July 2008 (UTC)[reply]

Well, when you put it that way, I can see where I should have been able to see it... Thanks! Dismas|^(talk) 00:11, 1 August 2008 (UTC)[reply]

Hope I'm not giving you a usless answer or making you confused, but there are cases where you multiply the exponents too. In (X^2)^3 would become X^(2*3) or X^6. --Wirbelwindヴィルヴェルヴィント (talk) 16:03, 1 August 2008 (UTC)[reply]

You can also regard (X^2)^3 as (X^2)(X^2)(X^2)=(XX)(XX)(XX)=XXXXXX=X^6 --Iamunknown 03:55, 3 August 2008 (UTC)[reply]