Re: Factor.....

*To*: mathgroup at smc.vnet.net*Subject*: [mg71092] Re: Factor.....*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Wed, 8 Nov 2006 06:14:26 -0500 (EST)*Organization*: The University of Western Australia*References*: <eicnmq$g3i$1@smc.vnet.net>

In article <eicnmq$g3i$1 at smc.vnet.net>, 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))) Seeking quadratic factors (using coercion into Series), rs = Solve[(a^2 + a c + b)(a^2 + a e + d) == (a^4 + y^2 + 3) + O[a]^5, {b, c, d, e}] we simplify the result using the fact that y is real. rs = Simplify[rs, Element[y, Reals]] One obtains 6 solutions as three conjugate pairs. Here are the three quadratic factorizations: (a^2 + a c + b)(a^2 + a e + d) /. rs // Union { (a^2 - I Sqrt[2] a (3 + y^2)^(1/4) - Sqrt[3 + y^2])* (a^2 + I Sqrt[2] a (3 + y^2)^(1/4) - Sqrt[3 + y^2]), (a^2 - I Sqrt[3 + y^2]) (a^2 + I Sqrt[3 + y^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]) } The last solution is the one you were looking for. > OR: > (y^2 - i·SQRT(A^4+3)) · (y^2 + i·SQRT(A^4+3)) You should have y not y^2 here. Seek quadratic factors of the form Solve[(y + b) (y + c) == (a^4 + y^2 + 3) + O[y]^3, {b, c}] Cheers, Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul