Algebraic manipulation

[Eddie W. Shore]

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