Bezout's Theorem is an important theorem in algebraic geometry. It is far
more general than what is presented here. It is, however, useful to
look at this specific case of the theorem when we have two curves in a
plane and want to know how they intersect.

**Definition. **The degree of a polynomial in two variables on the
form *x*^{r}*y*^{s} + *x*^{r-1}*y*^{s-1}
+ ... + *a* = 0 is max(r_{i} + s_{j}).

**Example. ***x*^{2}*y* +
*y* = 0 has degree
3.

**Definition.** The degree of a curve is equal to the degree of the
polynomial that defines the curve.

**Bezout's Theorem. **If we have two curves of degree r and s, they
will meet in exactly r*s points in **C**^{2} - the plane of
complex numbers.

**Example.** Let us consider the two curves given by the polynomial
equations *x*^{2} - *y *= 0 and *x + *2* *-
*y *= 0. The degrees of these polynomials are 2 and 1 respectively
and hence so are the degrees of the corresponding curves. Bezout's Theorem
tells us that the curves intersect in exactly 2 * 1 = 2 points.

Let us first do this algebraically. To find how the two curves meet,
we must set the two polynomials equal to each other and solve the resulting
equation.

*x*^{2} -
*y = x* + 2- *y*

*x*^{2} -
*x* - 2 = 0

A second order equation on the form
*ax*^{2 }+ *bx +
c = *0 has two solutions given by *x*_{1} = (-*b + *sqrt(*b*^{2
}- 4*ac*)) / 2*a *and *x*_{1} = (-*b + *sqrt(*b*^{2
}- 4*ac*)) / 2*a*.

We get *x*_{1} = 1/2* +
*sqrt(1^{ }- 4(-2))/2
= 1/2 + 3/2 = 2 and *x*_{2 }= -1. We get the points of intersection
by inserting these roots in one of the equations for the curves. For example
the linear equation gives us *y*_{1} = *x*_{1}
+ 2 = 2+ 2 = 4 and *y*_{2} = *x*_{2} + 2 = -1
+ 2 = 1.

We now plot the solutions to the two polynomial equations in Figure
1
and observe that the intersections are as predicted by our algebra and
the number of intersections are as predicted by Bezout's Theorem. The maximum
values for both axes are 5.