For which values of $k$ can $( x +y +z)^2 + k(x^2 +y^2 +z^2)$ be resolved into linear rational factors?

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My first attempt : Tried to solve by polynomial formulas but can't proceed after few steps.
Second attempt : Tried by vectors but found nothing useful.




abstract-algebra polynomials quadratics factoring




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    My first attempt : Tried to solve by polynomial formulas but can't proceed after few steps.
    Second attempt : Tried by vectors but found nothing useful.




    abstract-algebra polynomials quadratics factoring




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      My first attempt : Tried to solve by polynomial formulas but can't proceed after few steps.
      Second attempt : Tried by vectors but found nothing useful.




      abstract-algebra polynomials quadratics factoring




      share|cite|improve this question













      My first attempt : Tried to solve by polynomial formulas but can't proceed after few steps.
      Second attempt : Tried by vectors but found nothing useful.





      abstract-algebra polynomials quadratics factoring





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      edited Jul 28 at 17:34









      Batominovski

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      asked Jul 28 at 7:07









      user580093

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          If $ax^2+by^2+cz^2+2ryz+2sxz+2txy$ is a homogeneous quadratic
          in three variables, then it can only
          split into linear factors if the determinant
          $$beginvmatrixa&r&s\r&b&t\s&t&cendvmatrix=0.$$
          Here, that determinant is
          $$beginvmatrixk+1&1&1\1&k+1&1\1&1&k+1endvmatrix.$$






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            If $ax^2+by^2+cz^2+2ryz+2sxz+2txy$ is a homogeneous quadratic
            in three variables, then it can only
            split into linear factors if the determinant
            $$beginvmatrixa&r&s\r&b&t\s&t&cendvmatrix=0.$$
            Here, that determinant is
            $$beginvmatrixk+1&1&1\1&k+1&1\1&1&k+1endvmatrix.$$






            share|cite|improve this answer

























              up vote
              0
              down vote













              If $ax^2+by^2+cz^2+2ryz+2sxz+2txy$ is a homogeneous quadratic
              in three variables, then it can only
              split into linear factors if the determinant
              $$beginvmatrixa&r&s\r&b&t\s&t&cendvmatrix=0.$$
              Here, that determinant is
              $$beginvmatrixk+1&1&1\1&k+1&1\1&1&k+1endvmatrix.$$






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                If $ax^2+by^2+cz^2+2ryz+2sxz+2txy$ is a homogeneous quadratic
                in three variables, then it can only
                split into linear factors if the determinant
                $$beginvmatrixa&r&s\r&b&t\s&t&cendvmatrix=0.$$
                Here, that determinant is
                $$beginvmatrixk+1&1&1\1&k+1&1\1&1&k+1endvmatrix.$$






                share|cite|improve this answer













                If $ax^2+by^2+cz^2+2ryz+2sxz+2txy$ is a homogeneous quadratic
                in three variables, then it can only
                split into linear factors if the determinant
                $$beginvmatrixa&r&s\r&b&t\s&t&cendvmatrix=0.$$
                Here, that determinant is
                $$beginvmatrixk+1&1&1\1&k+1&1\1&1&k+1endvmatrix.$$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 28 at 7:14









                Lord Shark the Unknown

                84.6k950111




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