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)?

The polynomial $ \ Cx^2 + Bx + A \ $ is referred to as the reciprocal polynomial to $ \ Ax^2 + Bx + C \ \ , $ as its zeroes are the reciprocals of the latter polynomial.  In the same vein, a polynomial $ \ Ax^2 + Bx + A \ $  is called palindromic and has the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is as well.  (This applies to polynomials of higher degree as well as quadratics.)
This permits us to write
$$ ab·x^2 \ - \ (a^2+b^2)·x \ + \ ab \ \ = \ \ ab · \left[ \ x^2 \ - \ \frac{a^2+b^2}{ab}·x \ + \ 1 \ \right] $$ $$ = \ \ ab · \left[ \ x^2 \ - \ \left(\frac{b}{a } + \frac{a}{b}\right)·x \ + \ 1 \ \right] $$ $$ = \ \ ab·(x - r)· \left(x - \frac{1}{r} \right) \ \ = \ \ ab·[x^2 -  \left(r + \frac{1}{r} \right)·x \ + 1] \ \ . $$
The reciprocal nature of the two zeroes is thus confirmed and the factorization proceeds as in Taladris's answer.
