manipulating output of a quadratic expression after factor to a general form

First of all, welcome to the Maxima community!

Re "why this is not easily possible".  There are many ways to represent
expressions.  Factor chooses a canonical one, namely a rational function
(polynomial divided by polynomial).  After all, (x+1)/x = x*(1+1/x^2) =
(1/x)*(x^2+1) = 1 + 1/x etc.  An even simpler case: x^3/y^2 = x*(x/y)^2 =
(x^(3/2)/y)^2 etc. And for that matter, if it didn't choose a canonical
form, why wouldn't it return e.g. (x+1)*(1+1/x)/x instead of (x+1)^2/x^2 or
(1+1/x)^2?

In fact, for your example, Maxima doesn't even "think
of" (2*a*x+b)^2/(4*a^2) as a quotient, but as a product of three
multiplicands: 1/4, a^-2, and (2*a*x+b)^2.

That said, it certainly *is* possible to transform from one form to the
other by explicit (programmatic) transformations....

          -s


On Wed, Nov 16, 2011 at 15:49, <horatio at familiedeskranichs.info> wrote:

> I have this worksheet:
>
> Maxima 5.24.0 http://maxima.sourceforge.net
> using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (a.k.a. GCL)
> Distributed under the GNU Public License. See the file COPYING.
> Dedicated to the memory of William Schelter.
> The function bug_report() provides bug reporting information.
> (%i1) a * x^2 + b * x + c = 0;
>                                  2
> (%o1)                         a x  + b x + c = 0
> (%i2) % - c;
>                                   2
> (%o2)                          a x  + b x = - c
> (%i3) % / a;
>                                   2
>                                     a x  + b x     c
> (%o3)                          ---------- = - -
>                                            a          a
> (%i4) expand(lhs(%)) = rhs(%);
>                                       2   b x     c
> (%o4)                           x  + --- = - -
>                                             a      a
> (%i5) % + b^2/(2*a)^2;
>                                          2              2
>                                 2   b x    b         b     c
> (%o5)                     x  + --- + ---- = ---- - -
>                                     a        2          2   a
>                                           4 a       4 a
> (%i6) factor(lhs(%)) = rhs(%);
>                                                 2     2
>                                   (2 a x + b)     b      c
> (%o6)                       ------------ = ---- - -
>                                          2            2     a
>                                        4 a        4 a
> (%i7)
>
>
> but I would like to have the last expression as:
>
> (%i7) (x + b/(2*a))^2 = rhs(%);
>
>
>                                                      2
>
>
>                                          b  2    b     c
>
>
> (%o7)                        (x + ---)  = ---- - -
>
>
>                                         2 a        2   a
>
>
>                                                 4 a
>
>
> (%i8)
>
> So, if anyone knows a simple solution, or why this is not easily possible,
> please share.
> Thank you.
>
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