Re: Re: Factor.....

• To: mathgroup at smc.vnet.net
• Subject: [mg71110] Re: [mg71025] Re: Factor.....
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 8 Nov 2006 06:16:22 -0500 (EST)
• References: <eicnmq\$g3i\$1@smc.vnet.net><eieq1b\$l7h\$1@smc.vnet.net> <200611060752.CAA08527@smc.vnet.net>

```In fact, there is no need to define any new function, for this
particular problem anyway, since you can do this  as easily using the
already existing ones (and you get more answers as a bonus ;-))

(A^2 + a1*A + b1)*(A^2 + a2*A + b2) //.
{ToRules[Reduce[LogicalExpand[A^4 + 3 + y^2 == (A^2 + a1*A + b1)*(A^2
+ a2*A + b2) + O[A]^4],
{a1, a2, b1, b2}]]}

{(A^2 - Sqrt[-y^2 - 3])*(A^2 + Sqrt[-y^2 - 3]), (A^2 - Sqrt[-y^2 - 3])
*(A^2 + Sqrt[-y^2 - 3]),
(A^2 - Sqrt[2]*(y^2 + 3)^(1/4)*A + Sqrt[y^2 + 3])*(A^2 + Sqrt[2]*(y^2
+ 3)^(1/4)*A + Sqrt[y^2 + 3]),
(A^2 - I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3])*(A^2 + I*Sqrt[2]*
(y^2 + 3)^(1/4)*A -
Sqrt[y^2 + 3]), (A^2 - I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3])*
(A^2 + I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3]),
(A^2 - Sqrt[2]*(y^2 + 3)^(1/4)*A + Sqrt[y^2 + 3])*(A^2 + Sqrt[2]*(y^2
+ 3)^(1/4)*A + Sqrt[y^2 + 3])}

ExpandAll[%]

{A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 +
y^2 + 3, A^4 + y^2 + 3}

Andrzej Kozlowski
Tokyo, Japan

On 6 Nov 2006, at 16:52, ab_def at prontomail.com wrote:

> Along the same lines, we can define an extended SolveAlways function
> taking a third argument that specifies which parameters to solve for:
>
> mySolveAlways[\$Leq_, \$Lvar_, \$Lpar_ : {}] := Module[
>   {Leq = \$Leq, Lvar = \$Lvar, Lpar = \$Lpar, ans},
>   {Leq, Lvar, Lpar} = If[ListQ@ #, #, {#}]& /@
>     {Leq, Lvar, Lpar};
>   ans = Solve[!Eliminate[!And @@ Leq, Lvar], Lpar];
>   Select[ans, FreeQ[#, Alternatives @@ Lvar]&] /;
> ]
>
> In[2]:= (A^2 + a1*A + b1)*(A^2 + a2*A + b2) /.
>   Last@ mySolveAlways[
>     A^4 + 3 + y^2 == (A^2 + a1*A + b1)*(A^2 + a2*A + b2),
>     A, {a1, b1, a2, b2}]
>
> Out[2]= (A^2 - Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])*(A^2 +
> Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])
>
> Maxim Rytin
> m.r at inbox.ru
>
> dh wrote:
>> Hi,
>>
>> you have to tell Mathematica what form you want.
>>
>> Assume e.g. that we want the form (A^2 + a1 A + a0)(A^2 + b1 A + b0).
>>
>> Then we expand, equate the coefficients of A and solve for
>> a0,a1,b0,b1:
>>
>> r1=CoefficientList[(A^2+A a1+a0)(A^2+b1 A+ b0)//Expand,A]
>>
>> r2=CoefficientList[A^4+3+y^2,A]
>>
>>
>> this gives several possible expansions.
>>
>>
>>
>> Daniel
>>
>>
>>
>> gtsavdar at auth.gr wrote:
>>
>>> How can i factor A^4 + 3 + y^2 (A,y reals) for example with
>>
>>> Mathematica.....?
>>
>>>
>>
>>>
>>
>>> (
>>
>>> In order to have:
>>
>>> (A^2 + SQRT(y^2+3) + A·SQRT(2*SQRT(y^2+3))) · (A^2 + SQRT(y^2+3) -
>>
>>> A·SQRT(2*SQRT(y^2+3)))
>>
>>> )
>>
>>>
>>
>>> (
>>
>>> OR:
>>
>>> (y^2 - i·SQRT(A^4+3)) · (y^2 + i·SQRT(A^4+3))
>>
>>> )
>>
>>>
>>
>>>
>>
>>> Thanks....
>>
>>>
>

```

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