Here #f(x,y) = 2 x^2 + y^2 + 3 x y + 6 x - 15#
First we will classify the kind of conic we are handling. To do that we compute the matrix
#H = ((f_{x x},f_{xy}),(f_{yx},f_{yy}))#
where #f_{x_1x_2} =(partial^2f)/(partial x_1 partial x_2)#
obtaining
#H = ((4, 3), (3, 2))# with characteristic polynomial
#p(lambda) = lambda^2-6lambda-1#
whose roots are #{ 3 - sqrt[10],3 + sqrt[10]}#
being real and with opposite singns, characterizing a hyperbola with structure
#1/2(p-p_0).H.(p-p_0)+c#
Now, solving for #p_0, c# the condition
#1/2(p-p_0).H.(p-p_0)+c-f(x,y)=0, forall {x,y} in RR^2#
#
{
(2 x_0^2 + 3 x_0 y_0 + y_0^2 + 15 + c = 0),
( 3 x_0 + 2 y_0 =0),
(4 x_0 + 3 y_0 +6= 0)
:}
#
we obtain #x_0=12,y_0=-18,c=21#
A possible factorization is
#1/2(p-p_0).H.(p-p_0) = (c_1x+c_2 y+c_3)(c_4x+c_5y+c_6)#
giving
#f(x,y)=(x + y/2-3)(2 x + 2 y+12)+21#
Attached a plot with the hyperbola and the factorized lines.
