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Is it possible to write every real polynomial in two variables $x$ and $y$ with this form:

$$a x^2+b x y +c y^2$$

with general coefficients $a,b,c$ into the form

$$(d x + e y)^2$$

for some, possibly complex, $d$ and $e$?

From the second form to the first is obvious, but I need the other way around: is there some formula for the coefficients $d$ and $e$ as a function of $a,b,c$?

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    $\begingroup$ How about $x^2+y^2$? $\endgroup$ Commented Jul 10, 2020 at 14:50

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No, this does not happen for most polynomials. You can just divide the first expression thru by $y^2$ and obtain the expression $$a \left(\frac{x}{y}\right)^2 + b \left(\frac{x}{y}\right) + c$$ which is just a single-variable quadratic in $u = x/y$. It is well-known that the Fundamental Theorem of Algebra guarantees a factorization of this quadratic of the form $$a (u - r_1) (u - r_2)$$ but the roots can be different, so it may not be a perfect square. To recover the factorization for your original expression, just multiply thru by $y^2$ and you will see that it factors into $$a(x - r_1 y)(x - r_2 y) = (x\sqrt{a} - r_1 y\sqrt{a})(x\sqrt{a} - r_2 y\sqrt{a})$$ And hence you can see that such a factorization exists if and only if $r_1 = r_2$, which is equivalent to asserting that the discriminant $b^2 - 4ac = 0$.

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No it is not possible, $a = d^2$ and $c = e^2$. If this is true, then $d$ and $e$ must be real because a complex number of the form $t+ui$ where $u \neq 0$ is always going to be complex (verify this yourself). Thus, there are three constraints on two variables meaning that not every real polynomial will be spanned by this form even if $d$ and $e$ can be complex (because if they have imaginary parts the polynomial itself wont be a real polynomial anymore).

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