Find factors of this expression: $x^2-6xy+y^2+3x-3y+4$ 
Find factors of
  $$x^2-6xy+y^2+3x-3y+4.$$

I could not find the factors of this polynomial. Can you please explain me how to proceed?
 A: Hint: try with
$$ (x+ay+b)(x+cy+d)= x^2-6xy+y^2+3x-3y+4$$
Since this should be true for all $x,y$ you can put different values of $x,y$ into to calculate $a,b,c,d$. You should get a system of 4 equation.
Say for $y=0$ we get $$(x+b)(x+d)= x^2+3x+4$$
which should be true for all (real) $x$, but this is imposible since the discrimanant is $-7$. So you can't factor this expression (in real).
A: You cannot factor $f(x,y):=x^2-6xy+y^2+3x-3y+4$ over $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$.  If the base field is $\mathbb{Q}$ or $\mathbb{R}$, then the answer by greedoid covers it.  We now assume that the base field is $\mathbb{C}$ (or any algebraically closed field of characteristic $0$).
To show that $f(x,y)$ is an irreducible element of $\mathbb{C}[x,y]$, you can first homogenize the polynomial with a dummy variable $z$ to get
$$F(x,y,z)=x^2-6xy+y^2+3xz-3yz+4z^2\in\mathbb{C}[x,y,z]\,.$$
Then,
$$\frac{\partial F}{\partial x}(x,y,z)=2x-6y+3z\,,$$
$$\frac{\partial F}{\partial y}(x,y,z)=-6x+2y-3z\,,$$
and
$$\frac{\partial F}{\partial z}(x,y,z)=3x-3y+8z\,.$$
Thus, $\dfrac{\partial F}{\partial x}(x,y,z)=0$, $\dfrac{\partial F}{\partial y}(x,y,z)=0$, and $\dfrac{\partial F}{\partial z}(x,y,z)=0$ simultaneously if and only if $(x,y,z)=(0,0,0)$.  From this link, we conclude that $F(x,y,z)$ is irreducible over $\mathbb{C}$, and so is $f(x,y)$.

It can be shown that $f(x,y)$ is reducible over a given base field $K$ if and only if $\text{char}(K)=2$ or $\text{char}(K)=23$.  When $\text{char}(K)=2$, we have
$$f(x,y)=(x+y)\,(x+y+1)\,.$$
When $\text{char}(K)=23$, we have
$$f(x,y)=(7x+y+8)(10x+y+12)\,.$$
A: Compute the determinant
$$
\det\begin{bmatrix}
1 & -3 & 3/2 \\
-3 & 1 & -3/2 \\
3/2 & -3/2 & 4
\end{bmatrix}=-23
$$
Since the determinant is nonzero, the polynomial represents a nondegenerate conic, so it is irreducible.
A: Hint:
Use the quadratic formula and treat $x$ as a constant. What you will get is a function with asymptotic behaviour, i.e., something like $xy=1$.
