HP Prime polynomial long division bluesun08 Member Posts: 207 Threads: 48 Joined: Feb 2010 10-29-2013, 08:11 AM If i calculate (x^3+x):(x^2+3) manually via polynomial long division the result is x-(2x)/(x^2+3) How can i get this result with the HP Prime? Eddie W. Shore Posting Freak Posts: 764 Threads: 118 Joined: Aug 2007 10-29-2013, 09:02 AM Quote: If i calculate (x^3+x):(x^2+3) manually via polynomial long division the result is x-(2x)/(x^2+3) How can i get this result with the HP Prime? In CAS mode, use partfrac. Access: Toolbox, CAS, 1. Algebra, 6. Partial Fraction Hope that helps. bluesun08 Member Posts: 207 Threads: 48 Joined: Feb 2010 10-29-2013, 09:16 AM Great, thank you! bluesun08 Member Posts: 207 Threads: 48 Joined: Feb 2010 10-29-2013, 09:26 AM A further question: What is the difference between partfrac and cpartfrac? Eddie W. Shore Posting Freak Posts: 764 Threads: 118 Joined: Aug 2007 10-29-2013, 09:30 AM Quote: A further question: What is the difference between partfrac and cpartfrac? cpartfrac is the complex counterpart of partfrac Example: cpartfrac((x^3+x)/(x^2+3)) returns x - 1/(x+i*sqrt(3)) - 1/(x-i*sqrt(3)) bluesun08 Member Posts: 207 Threads: 48 Joined: Feb 2010 10-29-2013, 09:33 AM o.k. CR Haeger Junior Member Posts: 37 Threads: 7 Joined: Sep 2013 10-29-2013, 11:56 AM Thanks. Does CAS Settings / Simplify need to be set to NONE or MINIMUM for this result? Appears that MAXIMUM will show the rational being "recomposed" as x^3+x : x^2+3. Best Tim Wessman Posting Freak Posts: 1,278 Threads: 44 Joined: Jul 2007 10-29-2013, 12:18 PM Yes, it will recombine them in your version unforuntately. Note that this will all work much better in the future. There is a reason we did not put it to anything but "none" by default in the initial release. :-) TW Michael de Estrada Posting Freak Posts: 1,665 Threads: 142 Joined: Jan 2009 10-29-2013, 01:49 PM Hi Tim, A related issue. I want to expand and then recombine a symbolic polynomial to group the coefficients by powers, i.e. (a-x)*(b-x) ---> a*b - (a+b)*x + x^2 If I enter this expression in CAS with simplify set to maximum in the settings the result I get is: a*b - a*x - b*x + x^2 where the coefficients are not grouped for the power of x^1. Is there some way to do this ? Wes Loewer Junior Member Posts: 25 Threads: 1 Joined: Sep 2011 10-29-2013, 03:50 PM Quote: In CAS mode, use partfrac. To see the results of polynomial long division, propfrac is more what he was looking for. partfrac(((x^3+2*x²+3*x+4)/(x²-9))) returns the partial fractions x+2+(29/3/(x-3))+7/3/(x+3) while propfrac(((x^3+2*x²+3*x+4)/(x²-9))) returns the proper fraction: x+2+(12*x+22)/(x²-9) -wes Michael de Estrada Posting Freak Posts: 1,665 Threads: 142 Joined: Jan 2009 10-29-2013, 04:28 PM I found a solution in the CAS function "symb2poly". symb2poly ((a-x)*(b-x)) ---> [1 -a-b a*b] CompSystems Member Posts: 213 Threads: 60 Joined: Mar 2013 10-29-2013, 05:12 PM real collect Poly:= (a-x)*(b-x) Var:= x sum( coeff( Poly, Var ) .* seq( Var^k, k, degree( Poly, Var ), 0, 1 ) bluesun08 Member Posts: 207 Threads: 48 Joined: Feb 2010 10-29-2013, 05:14 PM "propfrac" is exactly the command i need. It's too bad that "propfrac" is not in CAS-Menü. parisse Junior Member Posts: 38 Threads: 0 Joined: Aug 2013 10-30-2013, 03:29 AM quo, rem and quorem are the CAS instructions to perform polynomial division. « Next Oldest | Next Newest »

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