Homogenous polynomials in $K[X,Y]$ split into linear factors [duplicate]

My question is quite simple.

Let $k$ be an algebraically closed field and $f\in k[X,Y]$. We know that $f$ can factor into linear polynomials. I would like to know if there is some generalization of this fact to $n$ indeterminates with $n\ge 3$.

Thanks in advance

EDIT. I should have said $f\in k[X,Y]$ a homogeneous polynomial, then $f$ can be factored into linear polynomials, is that true? This fact happens for polynomials in $k[X,Y,Z]$?

To a homogeneous polynomial $f(x,y)$ of degree $n$ we associate a polynomial $g(x)$ such that $f(x,y)=y^ng(x/y)$. Since $k$ is algebraically closed $g$ splits (in linear factors) in $k[x]$, so $g(x)=a(x-a_1)\cdots(x-a_n)$ and thus we get $f(x,y)=a(x-a_1y)\cdots(x-a_ny)$.

If $f\in k[x,y,z]$ then $f$ can be irreducible. For example, if the characteristic of $k$ is not $2$, then $x^2+y^2+z^2$ is irreducible. (For more details see $x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$.)

Basically, a homogeneous polynomial in three variables is the same as a polynomial in two variables. Collect all possible variables that appear in all the terms, with the highest possible exponent: you're left with $$ F(X,Y,Z)=X^a Y^b Z^c G(X,Y,Z) $$ where you can't collect any variable from $G$ (some or even all of $a,b,c$ can be zero). You have transferred the problem to $G(X,Y,Z)$. If one of the variables doesn't appear in $G$, then the polynomial is homogeneous in two variables (unless it was a monomial to begin with), so the facts you already know about it apply.

Thus you can assume all variables appear in $G$. Then you can consider $$ g(x,y)=G(x,y,1) $$ and you can write $$ G(X,Y,Z)=Z^k g(X/Z,Y/Z) $$ for some exponent $k$. Thus you see that reducibility of $G$ is equivalent to reducibility of $g$. A factorization of $g$ translates into a factorization of $G$ and conversely.

Say, for instance, that $G(X,Y,Z)=X^3-X^2Z+Y^2Z$. Then $g(x,y)=x^3-x^2+y^2$ and $k=3$. Since $g$ is irreducible, also $G$ is irreducible.