Tapeworm Posted January 2, 2013 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]
caKus Posted January 2, 2013 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).
mathematic Posted January 3, 2013 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.
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