Re: Factor and/or Rules replacements

*To*: mathgroup at smc.vnet.net*Subject*: [mg104496] Re: [mg104443] Factor and/or Rules replacements*From*: Adriano Pascoletti <adriano.pascoletti at dimi.uniud.it>*Date*: Sun, 1 Nov 2009 17:56:49 -0500 (EST)*References*: <200911010856.DAA19674@smc.vnet.net>

A possible solution rewrites {x -> rhs} as x-rhs and apply Times Adriano Pascoletti In[1]:= sol = Solve[x^2 + b*x + c == 0, {x}]; Times @@ (sol /. {{x_ -> y_} :> x - y}) Out[2]= ((1/2)*(b - Sqrt[b^2 - 4*c]) + x)*((1/2)*(b + Sqrt[b^2 - 4*c]) + x) In[3]:= Expand[%] Out[3]= c + b*x + x^2 2009/11/1 yves <yves.dauphin at solvay.com> > Hello everybody, > > Here is the model problem: > I tried to obtain the factorisation of x^2+b x +c and expected to get > something like (x-1/2(-b+Sqrt[b^2-4 c))((x-1/2(-b-Sqrt[b^2-4 c)). > > So I tried > in:Factor[x^2+b x + c] and get > out: c + b x + x^2 > So my first question what should I have done? > > I tried another way > in:sol=Solve[x^2+b x + c == 0,{x}]; and get the couple of replacement > rules. > Next I defined > in: f=(x-x1)(x-x2) > and finally I replaced x1 and x2 in succession by > in:f /. {(sol[[1]] /. x -> x1), (sol[[2]] /. x -> x2)} > which gave the expected factorisation but enclosed in {}. > My second question is : > Is there a simpler way or more compact to replace x1 and x2? > > With anticipated thanks. > Yves > >

**References**:**Factor and/or Rules replacements***From:*yves <yves.dauphin@solvay.com>