Algebraic manipulation
02-07-2019, 08:12 PM
Post: #1
 47box Junior Member Posts: 2 Joined: Jan 2019
Algebraic manipulation
Hi,

How can I manipulate a algebraic expression trying to extract a used defined term.

Exemple:

(4x-12) --> I know I can factor this term. Does exists any kind of control that I can specify a term?
I would like to specif "(x-3)" --> 4.(x-3)

Another Exemple:

(3-(4/5x)-(-5) --> I would like to specify a term "(x-10)" and see it´s possible.

Does the calculator can do that? Or any alternative?

Thanks
02-08-2019, 01:04 AM
Post: #2
 BruceH Senior Member Posts: 390 Joined: Dec 2013
RE: Algebraic manipulation
I don't know of a way to use a specific factor.

Alternatives:
• Use the factors() command to get a list of the factors. You can then inspect the list to see if the one you are expecting is present.
• Or, do a test divide by your factor, e.g. simplify((4*x-12)/(x-3))? If your divisor remains unchanged then it isn't a factor.
02-08-2019, 09:52 AM
Post: #3
 Aries Member Posts: 158 Joined: Oct 2014
RE: Algebraic manipulation
I think "factor()" and "simplify" will do the trick
Oh … and make sure you're in the CAS "ambient" !
Best,

Aries
02-08-2019, 05:37 PM
Post: #4
 47box Junior Member Posts: 2 Joined: Jan 2019
RE: Algebraic manipulation
Usually , I ´ve been using the factor and simplify command as mentioned. But I would like to specify a factor if possible...
Thanks

Best,

Mario
02-09-2019, 10:48 PM
Post: #5
 Eddie W. Shore Senior Member Posts: 1,327 Joined: Dec 2013
RE: Algebraic manipulation
Bruce,

I like your strategy the best - use the simplify(p(x)/q(x)).

The CAS program specfactor uses this strategy to factor q(x) from p(x):

Code:
 #cas specfactor(p,q):= BEGIN LOCAL r; r:=simplify(p/q); IF numer(r)==p THEN RETURN r; ELSE RETURN "("+STRING(r)+")"+"*("+STRING(q)+")"; END; END; #end

Successful factoring returns the result as strings (to preserve the format without automatic simplification):

specfactor(4x-12,x-3) returns "(4)*(x-3)" [ 4(x-3) ]

specfactor(8-4/5*x,x-10) returns "(-0.8)*(x-10)"

specfactor(x^3 -x^2+2*x-2,x^2+2) returns "(x-1)*(x^2+2)"

Eddie
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