f(x,y) = x^2 + 6xy + 9y^2 - 21y + 12
If f(x,y) is factorable, that implies in multiple tracing of f(x,y)=0
So, if f(x,y) is factorable for instance into f_1(x,y)f_2(x,y) this means that
f(x,y)=f_1(x,y)f_2(x,y)=0 has a minimum of two null leafs. One for f_1(x,y)=0 and the other for f_2(x,y)=0
In our case f(x,y) represents a slanted parabola having one null leaf, so this f(x,y) is not factorable.
We know that f(x,y) represents a parabola because putting it in the form
f(x,y)=1/2(x,y)cdot M cdot((x),(y))+(x,y)cdot b + c
M=((2, 6),(6, 18)) has characteristic polynomial
p_M(s) = s^2-20s=s(s-20)
which has a null root.
Attached the null trace of f(x,y)=0
