#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#
