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LOPANASTHĀPANĀBHYĀM
Lopana sthapanabhyam means 'by alternate elimination and retention'.
Consider the case of factorization of quadratic equation of type ax2 + by2 + cz2 + dxy + eyz + fzx This is a homogeneous equation of second degree in three variables x, y, z. The sub-sutra removes the difficulty and makes the factorization simple. The steps are as follows:
i) Eliminate z by putting z = 0 and retain x and y and factorize thus obtained a quadratic in x and y by means of ‘adyamadyena’ sutra.;

ii) Similarly eliminate y and retain x and z and factorize the quadratic in x and z.

iii) With these two sets of factors, fill in the gaps caused by the elimination process of z and y respectively. This gives actual factors of the expression.
Example 1:  3x2 + 7xy + 2y2 + 11xz + 7yz + 6z2.
Step (i):   Eliminate z and retain x, y; factorize
                3x2 + 7xy + 2y2 = (3x + y) (x + 2y)

Step (ii):   Eliminate y and retain x, z; factorize
                3x2 + 11xz + 6z2 = (3x + 2z) (x + 3z)

Step (iii):   Fill the gaps, the given expression
                 = (3x + y + 2z) (x + 2y + 3z)
Example 2:   12x2 + 11xy + 2y2 - 13xz - 7yz + 3z2.
Step (i):   Eliminate z i.e., z = 0; factorize
                12x2 + 11xy + 2y2 = (3x + 2y) (4x + y)

Step (ii):  Eliminate y i.e., y = 0; factorize
              12x2 - 13xz + 3z2 = (4x -3z) (3x – z)

Step (iii):  Fill in the gaps; the given expression
                = (4x + y – 3z) (3x + 2y – z)
Example 3:   3x2+6y2+2z2+11xy+7yz+6xz+19x+22y+13z+20
Step (i): Eliminate y and z, retain x and independent term
              i.e., y = 0, z = 0 in the expression (E).
        Then E = 3x2 + 19x + 20 = (x + 5) (3x + 4)

Step (ii): Eliminate z and x, retain y and independent term
               i.e., z = 0, x = 0 in the expression.
        Then E = 6y2 + 22y + 20 = (2y + 4) (3y + 5)

Step (iii): Eliminate x and y, retain z and independent term
                i.e., x = 0, y = 0 in the expression.
        Then E = 2z2 + 13z + 20 = (z + 4) (2z + 5)

Step (iv):   The expression has the factors (think of independent terms)
                 = (3x + 2y + z + 4) (x + 3y + 2z + 5).
In this way either homogeneous equations of second degree or general equations of second degree in three variables can be very easily solved by applying ‘adyamadyena’ and ‘lopanasthapanabhyam’ sutras.
Solve the following expressions into factors by using appropriate sutras:

        1.    x2 + 2y2 + 3xy + 2xz + 3yz + z2.

        2.    3x2 + y2 - 4xy - yz - 2z2 - zx.

        3.    2p2 + 2q2 + 5pq + 2p – 5q - 12.

        4.    u2 + v2 – 4u + 6v – 12.

        5.    x2 - 2y2 + 3xy + 4x - y + 2.

        6.    3x2 + 4y2 + 7xy - 2xz - 3yz - z2 + 17x + 21y – z + 20.
Highest common factor:
To find the Highest Common Factor i.e. H.C.F. of algebraic expressions, the factorization method and process of continuous division are in practice in the conventional system. We now apply' Lopana - Sthapana' Sutra, the 'Sankalana vyavakalanakam' process and the 'Adyamadya' rule to find out the H.C.F in a more easy and elegant way.
Example 1:   Find the H.C.F. of x2 + 5x + 4 and x2 + 7x + 6.

      1.  Factorization method:

                x2 + 5x + 4 = (x + 4) (x + 1)
                x2 + 7x + 6 = (x + 6) (x + 1)
                H.C.F. is ( x + 1 ).

     2. Continuous division process.

         x2 + 5x + 4 ) x2 + 7x + 6 ( 1
                             x2 + 5x + 4
                          ___________
                                  2x + 2 ) x2 + 5x + 4 ( ½x
                                              x2 + x
                                             __________
                                                    4x + 4 ) 2x + 2 ( ½
                                                                2x + 2
                                                                ______
                                                                    0

            Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.
3. Lopana - Sthapana process i.e. elimination and retention or alternate destruction of the highest and the lowest powers is as below:
                   
                   i.e.,, (x + 1) is H.C.F

Example 2:   Find H.C.F. of 2x2 – x – 3 and 2x2 + x – 6

                 


Example 3:    x3 – 7x – 6 and x3 + 8x2 + 17x + 10.

        Now by Lopana - Sthapana and Sankalana – Vyavakalanabhyam

                


Example 4:   x3 + 6x2 + 5x – 12 and x3 + 8x2 + 19x + 12.

               


                                            (or)

                    
Example 5:    2x3 + x2 – 9 and x4 + 2x2 + 9.

        By Vedic sutras:

                Add:  (2x3 + x2 – 9) + (x4 + 2x2 + 9)
                         = x4 + 2x3 + 3x2.

                  ÷ x2 gives      x2 + 2x + 3 ------ (i)

       Subtract after multiplying the first by x and the second by 2.

            Thus   (2x4 + x3 – 9x) - (2x4 + 4x2 + 18)
                      = x3 - 4x2 – 9x – 18 ------ ( ii )

       Multiply (i) by x and subtract from (ii)

                x3 – 4x2 – 9x – 18 – (x3 + 2x2 + 3x)
                 = - 6x2 – 12x – 18

                 ÷ - 6 gives     x2 + 2x + 3.
Thus ( x2 + 2x + 3 ) is the H.C.F. of the given expressions.
Algebraic Proof:
Let P and Q be two expressions and H is their H.C.F. Let A and B the Quotients after their division by H.C.F.

               P                   Q
        i.e., __  =  A  and   __  =  B     which gives P = A.H and Q = B.H
               H                   H

        P + Q = AH + BH and P – Q = AH – BH
                 = (A+B).H                 = (A–B).H
Thus we can write P ± Q = (A ± B).H
Similarly MP = M.AH and NQ = N.BH gives MP ± NQ = H (MA ± NB)
This states that the H.C.F. of P and Q is also the H.C.F. of P±Q or MA±NB.
i.e. we have to select M and N in such a way that highest powers and lowest powers (or independent terms) are removed and H.C.F appears as we have seen in the examples.
Find the H.C.F. in each of the following cases using Vedic sutras:

        1.   x2 + 2x – 8, x2 – 6x + 8

        2.   x3 – 3x2 – 4x + 12, x3 – 7x2 + 16x - 12

        3.   x3 + 6x2 + 11x + 6, x3 – x2 - 10x - 8

        4.   6x4 – 11x3 + 16x2 – 22x + 8,
              6x4 – 11x3 –  8x2 + 22x – 8.

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