Let $\mathcal{P}$ be the set of all bivariate polynomials over the reals of degree at most two, that are non-negative outside the $L_2$ unit ball. Are all the polynomials in $\mathcal{P}$ one of the following forms: $$p(x,y) = (ax+by+c)^2 \text{ or}$$ $$p(x,y) = \frac{(x-a_1)^2}{b_1^2} + \frac{(y-a_2)^2}{b_2^2} - 1 ?$$
If $p$ can be factorized into two different (upto constant factors) linear polynomials, then it will necessarily be negative somewhere be outside the unit ball. So, $p$ must be the square of a linear polynomial or irreducible.
If all irreducible polynomials ($n=2$, $d=2$, over reals) have the second type above (or a scalar multiple of it), then we are done. Is that true?
Edit 1: Not all irreducible polynomials ($n=2$, $d=2$, over reals) are of the second type above. For example, $x^2 -y^2 + 1$ is irreducible. But my original question, about $\mathcal{P}$, still stands.
Edit 2: As pointed out by obscurans in the comment, these are not the only two kinds polynomials in $\mathcal{P}$. It also seems that a polynomial in $\mathcal{P}$ does not even have to be the sum of squares of finitely many linear polynomials plus a (possibly negative) constant. $500y^2+200x^2-400xy+x-10$ seems to be a polynomial that cannot be wriiten as sum of squares of linear polynomials plus a constant (I could not prove this though). But it's in $\mathcal{P}$.
Edit 3: As Gerry Myerson pointed out in his comment, $500y^2+200x^2-400xy+x-10$ is in fact a sum of squares of linear polynomials plus a (negative) constant. Now I wonder if all polynomials in $\mathcal{P}$ are of this form...