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: 3x^{2} + 7xy + 2y^{2} = (3x + y)(x + 2y) 

2.  Repeat the above step for all, but one variables.  In this case, assuming y=0 gives: 3x^{2} + 11xz + 6z^{2} = (3x + 2z) (x + 3z) 

3.  Fill in the gaps, according to coefficients 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! 
Eliminating z, we get: 12x^{2} + 11xy + 2y^{2} = (3x + 2y)(4x + y) Eliminating y, we get: 12x^{2}  13xz + 3z^{2} = (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) 
Eliminating both y and z, we get: 3x^{2} + 19x + 20 = (x + 5)(3x + 4) Eliminating x and z, we get: 6y^{2} + 22y + 20 = (2y + 4)(3y + 5) Eliminating x and y, we get: 2z^{2} + 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 fillin the gaps. 
Steps  HCF  
1.  Multiply both the equations with terms, such that the highest power of indeterminate are same, and the coefficients of the highest power of the indeterminate are also the same.  In this case, the highest power is 2, in both the equations as x^{2}. And both the coefficients are 1 So, no multiplication is required. 

2.  Subtract one from the other to eliminate the terms with the highest power.  In this case, (x^{2} + 5x + 4)  (x^{2} + 7x + 6) = 2x  2 

3.  Retain only the term that does not have any common coefficient, or even indeterminate. Represent such that the first coefficient is positive.  In this case, (2x  2) has 2 as common factor. So, dividing by 2, we get: (x + 1), which is the answer! 
Since the terms with highest power are same, we eliminate by subtraction: (2x^{2} – x – 3)  (2x^{2} + 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) 
Since the terms with highest power are same, we eliminate by subtraction: (x^{3} – 7x – 6)  (x^{3} + 8x^{2} + 17x + 10) = 8x^{2}  24x  16 Dividing by common factor, 8, we get: x^{2} + 3x + 2 Thus, the HCF is (x^{2} + 3x + 2) 
Since the terms with highest power are same, we eliminate by subtraction: (x^{3} + 6x^{2} + 5x – 12)  (x^{3} + 8x^{2} + 19x + 12) = 2x^{2}  14x  24 Dividing by common factor, 2, we get: x^{2} + 7x + 12 Thus, the HCF is (x^{2} + 7x + 12) 
Since the terms with highest power are not same, we multiply accordingly: Mutiplying first expression by x, we get: 2x^{4} + x^{3}  9x And, multiplying second expression by 2, we get: 2x^{4} + 4x^{2} + 18 Now, we subtract to eliminate (2x^{4} + x^{3}  9x)  (2x^{4} + 4x^{2} + 18) = x^{3}  4x^{2}  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: (2x^{3} + x^{2} – 9) + (x^{4} + 2x^{2} + 9) = x^{4} + 2x^{3} + 3x^{2} We use this, for elimination by subtraction For which, we multiply our expression with x, to get: x^{4}  4x^{3}  9x^{2}  18x Subtracting, we get: (x^{4}  4x^{3}  9x^{2}  18x)  (x^{4} + 2x^{3} + 3x^{2}) = 6x^{3}  12x^{2}  18x Dividing by 6x, we get: x^{2} + 2x + 3 Thus, the HCF is (x^{2} + 2x + 3) 
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) 
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