<s>[INST] Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
 [/INST] For the title question. The factorization is given by
$$
(ax-b)(bx-a)=abx^2-(a^2+b^2)x+ab.
$$
We can find it by multiplying out $(\rho_1x+\rho_2)(\rho_3x+\rho_4)$ and comparing coefficients. This is easier than computing a discriminant.
</s>