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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?

MITjanitor
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Sidharth Ghoshal
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3 Answers3

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Generally, if you have a bunch of equations in several variables and want to solve them simultaneously, you should not think of factoring as the basic method but rather elimination of variables, one by one. For a different example, take the circle $x^2+y^2=2$ and the ellipse $x^2-xy+y^2=1$. You expect four intersections, if you count them properly anyway. Subtract the equations and get the third relation $xy=1$, and substitute $y=1/x$ into (for instance) the first equation, which becomes $x^4-2x^2+1=0$ after clearing the fractions. This is $(x^2-1)^2=0$, as you see, and now, in the last step, you factor, to see that the points $(1,1)$ and $(-1,-1)$ are the only intersection points, but that they occur with multiplicity two.

Lubin
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1

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.

amWhy
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  • This is the method I had used earlier. I didn't find it appealing at the time since the problem becomes exponentially more difficult to solve as equation degree and variable count is increased... – Sidharth Ghoshal Mar 25 '13 at 16:41
  • Do you suppose there is some way to divide out the roots the way one does when factoring a normal 1-dimensional polynomial? – Sidharth Ghoshal Mar 25 '13 at 16:41
  • The best way, with this given example, is to use the quadratic formula to find the roots of the equation in $z$: there are two real roots, but they will involve a radical (square root sign): $z = 7 \pm \dfrac{\sqrt{285}}{2}$, so those will be the roots for $(c)$, but each will be repeated. – amWhy Mar 25 '13 at 16:47
  • Does this make sense? In this case, the "fourth degree" polynomial can be treated as a quadratic equation, which greatly simplifies the factoring/solution-finding! – amWhy Mar 25 '13 at 17:47
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If you have equations of the form $p(x, y) = 0$ where $p$ is a polynomial, factoring that one is certainly worthwhile, It will give you more equations, but simpler ones. So, if you have: $$ \begin{align*} (x + y) (x^2 - y^2) &= 0 \\ (x - y) (x^2 + y^2) &= 0 \end{align*} $$ you end up with $2 \times 2 = 4$ (pick one from each) systems of equations.

But if factoring polynomials in one variable isn't a walk in the park, with more variables it just gets worse.

vonbrand
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