simple application of Bezout's Theorem Let $f(x),g(x) \in \mathbb{C}[x_1,\cdots,x_n]$ be two irreducible homogeneous polynomials of degree $n,m$ respectively. Does Bezout's Theorem say that the system of equations $f(x)=0, g(x)=0$ has precisely $m \cdot n$ solutions? 
Now let $y(y_1,\cdots,y_n)$ be a new set of indeterminates. Then $f(x+y),g(x+y)$ are still homogeneous of degree $m,n$ respectively, albeit they lie in the rings $\mathbb{C}[x_1,\cdots,x_n,y_1,\cdots,y_n]$. Does Bezout's Theorem say that the system of equations $f(x+y)=0, g(x+y)=0$ has still $m \cdot n$ solutions?
 A: Edit:  As noted by Sanchez, condition (3) had to be corrected from its original formulation.
No, what is true is that if you take $n$ polynomials of degrees $d_1, d_2, \dots, d_n$, in $k[x_1,\dots,x_n]$ then they have exactly $d_1 d_2 \cdots d_n$ common solutions provided:
(1) you count your solutions over an algebraically closed field (no problem here, since you're working over $k = \mathbb{C}$);
(2) you count the number of solutions in projective space (i.e. you have to homogenize your system of equations and then only count nonzero solutions up to rescaling);
(3) the $n$ polynomials must intersect in only a finite number of points (it is necessary, but not sufficient, that the $n$ polynomials are pairwise relatively prime); this condition will hold for a generic choice of $n$ polynomials in $k[x_1,\dots,x_n]$;
(4) you count each solution with an appropriate intersection multiplicity.
In addition to these requirements, Bezout's theorem is only applicable when the number of equations is the same as the number of variables; intuitively, you are starting in an $n$-dimensional space $\mathbb{C}^n$ -- actually $\mathbb{CP}^n$ by (2) -- and each of the $n$ equations cuts down the common solution space by $1$ dimension, resulting in a $0$ dimension intersection (i.e. a finite set of points).  Condition (3) is to insure that you indeed get a $0$ dimensional intersection, while conditions (1) and (4) make sure that you count every solution, the intersection multiplicities coming in because you must count some solutions repeatedly as necessitated by the relevant algebra in the proof.
