A bivariate linear function has 3 coefficients (degrees of freedom) and (explicitly) describes a plane:
$z=f_1(x,y)=a_1+a_2x+a_3y$
A bivariate quadratic function has 6 coefficients (degrees of freedom) and (explicitly) describes a paraboloid.
$z=f_2(x,y)=b_1+b_2x+b_3y+b_4xy+b_5x^2+b_6y^2$
A bivariate cubic function has 10 coefficients (degrees of freedom).
$z=f_3(x,y)=c_1+c_2x+c_3y+c_4xy+c_5x^2+c_6y^2+c_7x^2y+c_8xy^2+c_9x^3+c_{10}y^3$
I presume the product $f_{2.5}(x,y)=f_1(x,y)f_2(x,y)$ has 3+6-1=8 degrees of freedom ("-1" arising from the fact that scaling $f_1$ and scaling $f_2$ have the same effect on $f_{2.5}$).
In summary:
DoF Shape
linear 3 z=f_1 plane
quadratic 6 z=f_2 paraboloid
linear x quadratic 8 z=f_2.5 ?
cubic 10 z=f_3
Question: Is there anything interesting we can say about the geometric nature of $f_{2.5}$? We have intuition on what a plane ($f_1$) looks like, and what paraboloids (${f_2}$) look like. What can we say (even in vague language) about what $f_{2.5}$ looks like?
Given $$\begin{aligned} f_1(x,y) &= a_1 + a_2 x + a_3 y \\ f_2(x,y) &= b_1 + b_2 x + b_3 y + b_4 x y + b_5 x^2 + b_6 y^2 \\ f_3(x,y) &= c_1 + c_2 x + c_3 y + c_4 x y + c_5 x^2 + c_6 y^2 + c_7 x^2 y + c_8 x y^2 + c_9 x^3 + c_{10} y^3 \\ \end{aligned}$$ then $$f_1(x, y) f_2(x, y) = f_3(x, y)$$ where $$\begin{aligned} c_1 &= a_1 b_1 \\ c_2 &= a_1 b_2 + a_2 b_1 \\ c_3 &= a_1 b_3 + a_3 b_1 \\ c_4 &= a_1 b_4 + a_2 b_3 + a_3 b_2 \\ c_5 &= a_1 b_5 + a_2 b_2 \\ c_6 &= a_1 b_6 + a_3 b_3 \\ c_7 &= a_3 b_5 + a_2 b_4 \\ c_8 &= a_2 b_6 + a_3 b_4 \\ c_9 &= a_2 b_5 \\ c_{10} &= a_3 b_6 \\ \end{aligned}$$
When you multiply a polynomial of degree $N$ with a polynomial of degree $M$, you get a polynomial of degree $N+M$. This applies even to multivariate polynomials.