 Algebraic manipulation - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Algebraic manipulation (/thread-12378.html) Algebraic manipulation - 47box - 02-07-2019 08:12 PM 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 RE: Algebraic manipulation - BruceH - 02-08-2019 01:04 AM 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. RE: Algebraic manipulation - Aries - 02-08-2019 09:52 AM I think "factor()" and "simplify" will do the trick Oh … and make sure you're in the CAS "ambient" ! Best, Aries RE: Algebraic manipulation - 47box - 02-08-2019 05:37 PM 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 RE: Algebraic manipulation - Eddie W. Shore - 02-09-2019 10:48 PM 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