Vtyjkyuk

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?

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).

Compute the determinant
$$
detbeginbmatrix
1 & -3 & 3/2 \
-3 & 1 & -3/2 \
3/2 & -3/2 & 4
endbmatrix=-23
$$
Since the determinant is nonzero, the polynomial represents a nondegenerate conic, so it is irreducible.

You cannot factor $f(x,y):=x^2-6xy+y^2+3x-3y+4$ over $mathbbQ$, $mathbbR$, or $mathbbC$. If the base field is $mathbbQ$ or $mathbbR$, then the answer by greedoid covers it. We now assume that the base field is $mathbbC$ (or any algebraically closed field of characteristic $0$).

To show that $f(x,y)$ is an irreducible element of $mathbbC[x,y]$, you can first homogenize the polynomial with a dummy variable $z$ to get
$$F(x,y,z)=x^2-6xy+y^2+3xz-3yz+4z^2inmathbbC[x,y,z],.$$
Then,
$$fracpartial Fpartial x(x,y,z)=2x-6y+3z,,$$
$$fracpartial Fpartial y(x,y,z)=-6x+2y-3z,,$$
and
$$fracpartial Fpartial z(x,y,z)=3x-3y+8z,.$$
Thus, $dfracpartial Fpartial x(x,y,z)=0$, $dfracpartial Fpartial y(x,y,z)=0$, and $dfracpartial Fpartial z(x,y,z)=0$ simultaneously if and only if $(x,y,z)=(0,0,0)$. From this link, we conclude that $F(x,y,z)$ is irreducible over $mathbbC$, 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 $textchar(K)=2$ or $textchar(K)=23$. When $textchar(K)=2$, we have
$$f(x,y)=(x+y),(x+y+1),.$$
When $textchar(K)=23$, we have
$$f(x,y)=(7x+y+8)(10x+y+12),.$$

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 $xy=1$.