factorize symbolic expression

asked 2019-12-15 15:36:29 +0100

Jingenbl
31 ●1 ●6

(ax+bx).factor()=x(a+b) but (2a+2b).factor()=(2a+2b) how to obtain 2(a+b) ?

Thanks in advance...

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There exists the collect_common_factors method that should do the trick, I think. But it seems a bit buggy. This works as expected:

sage: a, b = var("a, b")
sage: (2*pi*a + 2*pi*b).collect_common_factors()
2*pi*(a + b)
sage: (2*a^2 + 2*a*b).collect_common_factors()
2*(a + b)*a

However, this fails:

sage: (2*a + 2*b).collect_common_factors() 
2*a + 2*b
sage: (SR(2)*a + SR(2)*b).collect_common_factors()
 2*a + 2*b

Juanjo ( 2019-12-16 01:16:07 +0100 )edit

Looks like a bug to me, since (2*a*x + 2*b*x).factor() returns 2*(a + b)*x.

John Palmieri ( 2019-12-18 18:44:28 +0100 )edit

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answered 2019-12-17 22:04:43 +0100

John Palmieri

8876 ●19 ●73 ●192 http://www.math.washin...

You can work with elements of a polynomial ring instead of with symbolic expressions:

sage: R.<a,b,c> = ZZ[] # polynomials with integer coefficients, variables a, b, c
sage: (a*2+b*2).factor()
2 * (a + b)
sage: (a*c+b*c).factor()
c * (a + b)

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Thanks for the replay. A little bit heavy for a simple factorization of an expression... Any particular reason for this restriction of the format method? Big thanks anyway

Jingenbl ( 2019-12-18 10:33:25 +0100 )edit

The notion of factorization is algebraic and depends on the ring in which you are working. For example x^2+1 factors over the complex numbers but not the reals, and x^2-2 factors over the reals but not the rationals. So when you're dealing with factorization, it's a good idea to precisely specify the ring.

John Palmieri ( 2019-12-18 18:43:43 +0100 )edit

ok I undersatnd....

Jingenbl ( 2019-12-19 09:42:52 +0100 )edit

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Asked: 2019-12-15 15:36:29 +0100

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Last updated: Dec 17 '19

factorize symbolic expression edit

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