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I want to find a factorization of the following polynomial of degree 2.

Let $\alpha >0, \beta > 0,$ and $\gamma \geq 1$. Is there any hope to factorize the following polynomial: $$ \alpha^2 x^2 + \beta^2 y^2 - 2 \alpha \beta \gamma xy = C(Ax- By)^2, $$ for some positive constants $A, B, C$.

Thank you in advance

S

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  • $\begingroup$ I just edited your post. The word is 'polynomial' in English (in fact, you spelt it correctly in your first line!). $\endgroup$
    – Toby Mak
    Nov 15, 2019 at 10:52
  • $\begingroup$ Well done. Thank you! $\endgroup$
    – sidiatig
    Nov 15, 2019 at 10:55

2 Answers 2

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If we expand the RHS first:

$C(Ax-By)^2=C(A^2x^2+B^2y^2-2ABxy)=CA^2x^2+CB^2y^2-2ABCxy$

Comparing coefficients with the LHS now:

$CA^2=\alpha^2$ (1)

$CB^2=\beta^2$ (2)

$-2ABC=-2\alpha\beta\gamma\implies ABC=\alpha\beta\gamma$

Multiplying together (1) and (2), we have $C^2A^2B^2=\alpha^2\beta^2$, which means that your polynomial can only be factored into the form $C(Ax-By)^2$ if $\gamma=1$.

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  • $\begingroup$ Even more, I think it an "iff" $\gamma = 1$. Thank you for your suggestions. Actually I try to lower bound this polynomial. Please could you check this math.stackexchange.com/questions/3436633/… $\endgroup$
    – sidiatig
    Nov 15, 2019 at 11:15
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Old answer

Your expression can be factorized of we set $\gamma = \frac{1}{\sqrt{\alpha\beta}}$. In that case, $C=1, A=\sqrt{\alpha}$, and $B=\sqrt{\beta}$.

Edit: as done by JY1853, comparing coefficients is the way.

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  • $\begingroup$ Sorry I edit the polynom to $\alpha^2 x^2 + \beta^2 y^2 - 2 \alpha \beta \gamma xy$. When I did some identification using development I get the following: $$ A = \frac{\alpha \gamma}{\beta}, B = \gamma, C = \frac{\beta^2}{\gamma^2}$$ However developing $C(Ax- By)^2 = \alpha^2 x^2 + \beta^2 y^2 - 2\alpha\beta xy$ (there is no $\gamma$ term in front of the monom $xy$. :( $\endgroup$
    – sidiatig
    Nov 15, 2019 at 10:30

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