Tapeworm Posted January 2, 2013 Share Posted January 2, 2013 I don't know the procedure to factorize: [latex]a^2-ab+b^2-bc+c^2-ca[/latex] into [latex](a+\omega b+\omega^2 c)(a+\omega^2 b+\omega c)[/latex] [latex]\omega[/latex] is complex cube root of unity: [latex]\omega^3=1[/latex] ============================================= Can all the quadratic forms be factorized with complex roots? [latex]ax^2+2fxy+by^2+2gyz+2px+2qy+d=0[/latex] Link to comment Share on other sites More sharing options...
caKus Posted January 2, 2013 Share Posted January 2, 2013 Develop the product of the second form, group by similar term (a^2, ab, b^2 ...) then check that the coeficients of each similar term are equal in form (1) and (2). Link to comment Share on other sites More sharing options...
mathematic Posted January 3, 2013 Share Posted January 3, 2013 I don't know the procedure to factorize: [latex]a^2-ab+b^2-bc+c^2-ca[/latex] into [latex](a+\omega b+\omega^2 c)(a+\omega^2 b+\omega c)[/latex] [latex]\omega[/latex] is complex cube root of unity: [latex]\omega^3=1[/latex] ============================================= Can all the quadratic forms be factorized with complex roots? [latex]ax^2+2fxy+by^2+2gyz+2px+2qy+d=0[/latex] The trick is ω + ω² + 1 = 0. This gives the coefficients for ab, bc, and ac. Link to comment Share on other sites More sharing options...
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