Sylvester Resultant Formulation

The Sylvester resultant formulation only handles the case of two polynomials: f(x1,x2), g(x1,x2).
The sylvester resultant is based upon the following observation: we can symbolically multiply f by another polynomial h1, and multiply g by another polynomial h2 using linear algebra. Furthermore, if h1 has degree 1 less than g and h2 has degree one less than f, then f and g must share a common root. The reason we make the "degree 1 less than" requirement is that we are going to compute the difference f(x,y)h1(x,y) - g(x,y)h2(x,y) and assume it is equal to 0; but, this sum is only interesting if h1 has degree less than g and h2 has degree less than f because otherwise, we could simply let h1 be equal to (or a product of) g and h2 be equal to (or a product of) f, and achieve the condition that f(x,y)h1(x,y) - g(x,y)h2(x,y) is 0.
Consider the following two polynomials: f(x,y) = x^2 (6y^2 + 3y + 4) + x (5y^2 - 4y - 1) + (2y^2 + 3)
g(x,y) = x^2 (5y^2 + 7y + 5) + x (3y^2 - 9y - 7) + (y^2 - 4x - 2)
which we can rewrite as: f(x,y) = x^2 fx2 + x fx1 + fx0
g(x,y) = x^2 gx2 + x gx1 + gx0
Consider two functions h1 = x h1x1 + h1x0 and h2 = x h2x1 + h2x0, which are functions solely of x (not y). Furthermore, consider the difference of their products f(x,y)h1(x,y) - g(x,y)h2(x,y) which can be written in linear algebra form as
fx1(y) fx2(y) 0
fx0(y) 0 gx0(y) 0
fx0(y) gx1(y) gx0(y)
fx1(y) gx2(y) gx1(y)
fx2(y) 0 gx2(y)
h1x0 h1x1 -h2x0 -h2x1

If f(x',y')=g(x',y')=0, then, for some y, f(x') and g(x') share a common root, let's call alpha. Since f(x') and g(x') share a common root, then let h1=g/(x-alpha) and let h2=f/(x-alpha), and in this case, the linear algebra product f(x,y)h1(x,y) - g(x,y)h2(x,y) will be [0, 0, 0, 0].
Again, by Nullstellenstatz, if the linear algebra product is [0, 0, 0, 0], and since the column vector is non-zero, it must be the case that the determinant of the matrix is 0. Consequently, we have constructed a univariate polynomial (Det(Matrix(y))) whose roots (y') satisfy the property that any common root of (f(x,y),g(x,y)) is also a root of Q(y).
In the above example, the matrix would be (5y^2 - 4y - 1) (6y^2 + 3y + 4) 0
(2y^2 + 3) 0 (y^2 - 4x - 2) 0
(2y^2 + 3) (3y^2 - 9y - 7) (y^2 - 4x - 2)
(5y^2 - 4y - 1) (5y^2 + 7y + 5) (3y^2 - 9y - 7)
(6y^2 + 3y + 4) 0 (5y^2 + 7y + 5)