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उपसूत्र ११. लोपनस्थापनाभ्यां

(Upasūtra 11. lopanasthāpanābhyāṃ) - By alternate elimination and retention.

The Upasūtra: lopanasthāpanābhyāṃ (By alternate elimination and retention) is used for factorization of homogeneous Quadratic expressions for two or more variables.
 
As an illustration, let us use this Upasūtra, in simple steps, to factorize:
3x2 + 7xy + 2y2 + 11xz + 7yz + 6z2
  Steps Factors
1. Assume one of them as Zero, to eliminate and factorize the remaining expression (use Upasūtra: ādyamādyenāntyamantyena for it). In case of an independent term, eliminate more variables In this case, there is no independent term.
Assuming z=0 gives:
3x2 + 7xy + 2y2
= (3x + y)(x + 2y)
2. Repeat the above step for all, but one variables. In this case, assuming y=0 gives:
3x2 + 11xz + 6z2
= (3x + 2z) (x + 3z)
3. Fill in the gaps, according to co-efficients of the variable not eliminated (not assumed as Zero). In case of an independent term, fill the gaps in accordance to the independent terms. In this case,
(3x + 2z) and (3x + y) has 3x
So, we get: (3x + y + 2z)
 
Again, (x + 2y) and (x + 3z) has 1x
So, we get: (x + 2y + 3z)
 
Therefore the factors are:
(3x + y + 2z)(x + 2y + 3z), which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
12x2 + 11xy + 2y2 - 13xz - 7yz +3z2
  Eliminating z, we get: 12x2 + 11xy + 2y2
= (3x + 2y)(4x + y)
 
Eliminating y, we get: 12x2 - 13xz + 3z2
= (4x - 3z)(3x - z)
 
Filling the gaps, we get:
(3x + 2y - z)(4x + y - 3z)
 
Thus, the factors are:
(3x + 2y - z)(4x + y - 3z)
Again, for something like:
3x2 + 6y2 + 2z2 + 11xy + 7yz + 6xz + 19x + 22y + 13z + 20
  Eliminating both y and z, we get: 3x2 + 19x + 20
= (x + 5)(3x + 4)
 
Eliminating x and z, we get: 6y2 + 22y + 20
= (2y + 4)(3y + 5)
 
Eliminating x and y, we get: 2z2 + 13z + 20
= (z + 4)(2z + 5)
 
Filling the gaps, we get:
(3x + 2y + z + 4)(x + 3y + 2z + 5)
 
Thus, the factors are:
(3x + 2y + z + 4)(x + 3y + 2z + 5)
 
Note that, here we have an independent term. So, we elimiate two at a time and use the independent terms of the factors to fill-in the gaps.
 
 
Using Upasūtra 11. lopanasthāpanābhyāṃ for finding HCF:
It would seem that this Upasūtra would be helpful for HCF (Highest Common Factor) because of its ease. But, this Upasūtra is also discussed for HCF using a different technique, in addition to the factorization technique explained above.
 
As an illustration, let us use this Upasūtra, in simple steps, for HCF of:
x2 + 5x + 4, and
x2 + 7x + 6
  Steps HCF
1. Multiply both the equations with terms, such that the highest power of indeterminate are same, and the co-efficients of the highest power of the indeterminate are also the same. In this case, the highest power is 2, in both the equations as x2. And both the co-efficients are 1
So, no multiplication is required.
2. Subtract one from the other to eliminate the terms with the highest power. In this case,
(x2 + 5x + 4) - (x2 + 7x + 6)
= -2x - 2
3. Retain only the term that does not have any common co-efficient, or even indeterminate. Represent such that the first co-efficient is positive. In this case, (-2x - 2) has -2 as common factor.
So, dividing by -2, we get:
(x + 1), which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
2x2 – x – 3, and
2x2 + x – 6
  Since the terms with highest power are same, we eliminate by subtraction:
(2x2 – x – 3) - (2x2 + x – 6)
= -2x + 3
Dividing by common factor, -1, to make the first term positive, we get:
2x - 3
 
Thus, the HCF is (2x - 3)
Again, for something like:
x3 – 7x – 6, and
x3 + 8x2 + 17x + 10
  Since the terms with highest power are same, we eliminate by subtraction:
(x3 – 7x – 6) - (x3 + 8x2 + 17x + 10)
= -8x2 - 24x - 16
Dividing by common factor, -8, we get:
x2 + 3x + 2
 
Thus, the HCF is (x2 + 3x + 2)
And, for something like:
x3 + 6x2 + 5x – 12, and
x3 + 8x2 + 19x + 12
  Since the terms with highest power are same, we eliminate by subtraction:
(x3 + 6x2 + 5x – 12) - (x3 + 8x2 + 19x + 12)
= -2x2 - 14x - 24
Dividing by common factor, -2, we get:
x2 + 7x + 12
 
Thus, the HCF is (x2 + 7x + 12)
Finally, for something like:
2x3 + x2 – 9, and
x4 + 2x2 + 9
  Since the terms with highest power are not same, we multiply accordingly:
Mutiplying first expression by x, we get:
2x4 + x3 - 9x
And, multiplying second expression by 2, we get:
2x4 + 4x2 + 18
 
Now, we subtract to eliminate (2x4 + x3 - 9x) - (2x4 + 4x2 + 18)
= x3 - 4x2 - 9x - 18
 
But, this expression contains highest power of one of the expressions.
This will require us to further eliminate the highest power of indeterminate by using Sūtra 7. saṅkalana vyavakalanābhyāṃ
 
Addition of the expression, gives:
(2x3 + x2 – 9) + (x4 + 2x2 + 9)
= x4 + 2x3 + 3x2
We use this, for elimination by subtraction
For which, we multiply our expression with x, to get:
x4 - 4x3 - 9x2 - 18x
Subtracting, we get:
(x4 - 4x3 - 9x2 - 18x) - (x4 + 2x3 + 3x2)
= -6x3 - 12x2 - 18x
Dividing by -6x, we get:
x2 + 2x + 3
 
Thus, the HCF is (x2 + 2x + 3)
But, why does it work? For this Upasūtra (lopanasthāpanābhyāṃ), let us consider the following:
  Assuming two expressions, p and q, with HCF: H.
And, let A and B be the expressions of quotients that remains after dividing by their HCF
 
Then, p/H = A and q/H = B
Thus, p = AH and q = BH
 
Now, p + q = AH + BH = (A + B)H
And, p - q = AH - BH = (A - B)H
Thus, p ± q = (A ± B)H
 
Similarly, for any expression m and n, multiplied with p and q:
mp = mAH and nq = nBH, which gives:
mp ± nq = (mA ± nB)H
 
Thus, the choice of m and n in a manner that they help in eliminating the highest powers would lead to the HCF:
H = (HCF of p and q) = (HCF of p±q) = (HCF of mA± nB)
 
Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »
 
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