The factorization of a linear partial differential operator (LPDO) is an important issue in the theory of integrability, due to the Laplace-Darboux transformations, which allow construction of integrable LPDEs. Laplace solved the factorization problem for a bivariate hyperbolic operator of the second order (see Hyperbolic partial differential equation), constructing two Laplace invariants. Each Laplace invariant is an explicit polynomial condition of factorization; coefficients of this polynomial are explicit functions of the coefficients of the initial LPDO. The polynomial conditions of factorization are called invariants because they have the same form for equivalent (i.e. self-adjoint) operators. Beals-Kartashova-factorization (also called BK-factorization) is a constructive procedure to factorize a bivariate operator of the arbitrary order and arbitrary form. Correspondingly, the factorization conditions in this case also have polynomial form, are invariants and coincide with Laplace invariants for bivariate hyperbolic operators of the second order. The factorization procedure is purely algebraic, the number of possible factorizations depending on the number of simple roots of the Characteristic polynomial (also called symbol) of the initial LPDO and reduced LPDOs appearing at each factorization step. Below the factorization procedure is described for a bivariate operator of arbitrary form, of order 2 and 3. Explicit factorization formulas for an operator of the order can be found in General invariants are defined in and invariant formulation of the Beals-Kartashova factorization is given in Beals-Kartashova Factorization Operator of order 2 Consider an operator with smooth coefficients and look for a factorization Let us write down the equations on explicitly, keeping in mind the rule of left composition, i.e. that Then in all cases where the notation is used. Without loss of generality, i.e. and it can be taken as 1, Now solution of the system of 6 equations on the variables can be found in three steps. At the first step, the roots of a quadratic polynomial have to be found. At the second step, a linear system of two algebraic equations has to be solved. At the third step, one algebraic condition has to be checked. Step 1. Variables can be found from the first three equations, The (possible) solutions are then the functions of the roots of a quadratic polynomial: Let be a root of the polynomial then Step 2. Substitution of the results obtained at the first step, into the next two equations yields linear system of two algebraic equations: In particularly, if the root is simple,