Nullified terms of this polynomial?

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This the polynomial:

$6 + 3x - 5y + 7xy -8x^2 + 3y^2 + 4x^2y - 7xy^2 + 9x^3 - 5x^4 + 8y^4x$

I would like to know which terms are there (so that I will form the complete polynomial, with zero as a coefficient for these terms).

For example, in this polynomial:

$-5 + y + 9x + yx^2$

I know that the "complete" polynomial is:

$-5 + y + 9x + 0xy + 0x^2 + yx^2$

Notice how the "complete" polynomial bounds the "hidden" terms by its "max" term.

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8
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If there are terms $x^i y^j$ where $i \le n_x, j \le n_y$ then there are a total of $(n_x+1)(n_y+1)$ terms.

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copper.hat's answer is very reasonable (and in particular matches your example of total degree 3).

But be careful - there are other, very sound systems. The "true" answer is: "it depends". More precisely, which terms are missing depends on the monomial order of your underlying polynomial ring, which for many applications is irrelevant and, thus, unspecified. Also, for some (many) standard monomial orders, it could well be infinitely many terms.

If I want to stress the fact that the polynomial is, say, quadratic in $x$ and linear in $y$, like in your template, I'd use the $i\le n_x, j \le n_y$ rule as proposed, but also order the terms and parenthesise as $$x^2 (a_{21}y+a_{20}) + x (a_{11}y + a_{10}) + a_{00}$$ for clarity and to make sure that my intentions come across.

Otherwise -- and in particular for a rather dense polynomial -- my natural choice would be not to use any terms where the total degree of the monomial exceeds the total degree of the original polynomial, and I'd order the terms by total degree and, in case of ties, lexicographically. In your example, the total degree is 5, and I would fill in all $x^iy^j$ with $i+j \le 5$, as in copper.hat's comment - that's only 21 instead of 25 terms.