Factoring 3 Dimensional Polynomials? How do you factor a system of polynomials into their roots the way one can factor a single dimensional polynomial into its roots.
Example
$$x^2 + y^2 = 14$$
$$xy = 1$$
We note that we can find the 4 solutions via quadratic formula and substitution such that the solutions can be separated into $2$ groups of $2$ such that each group lies on either $(x + y + 4)$ and $(x + y - 4)$. Note that:
$$(x + y + 4)(x + y - 4) = 0$$
$$xy = 1$$
Is also an equivalent system.
How do I factor the bottom half?
Ideally if $g$ is a linear expression then my system should be
$$g_1 * g_2 = 0$$
$$g_3 * g_4 = 0$$
Such that the solutions to
Any of the subsystems of this system is a solution to system itself (note there are $4$ viable subsystems).
Help?

For the "example" you list, here are some suggestions: Given 
$$x^2 + y^2 = 14\tag{1}$$
$$xy = 1 \iff y = \frac 1x\tag{2}$$


*

*Substitute $y = \dfrac{1}{x}\tag{*}$ into equation $(1)$. Then solve for roots of the resulting equation in one variable.


$$x^2 + y^2 = 14\tag{a}$$
$$\iff x^2 + \left(\frac 1x\right)^2 = 14\tag{b}$$
$$\implies x^4 - 14x^2 + 1 = 0\tag{c}$$
Try letting $z = x^2$ and solve the resulting quadratic:
$$z^2 - 14z + 1 = 0\tag{d}$$ Then for each root $z_1, z_2,\;$ there will be two repeated roots solving $(c)$: $2 \times 2$ repeated roots
Notice that by incorporating the equation (2) into equation  step $(b)$, we effectively end up finding all solutions that satisfy both equations $(1), (2)$: solutions for which the two given equations coincide.
See Lubin's post for some additional "generalizations": ways to approach problems such as this.
