<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] Using the Quadratic Formula is more general but here is an alternative method:
We have $abx^2-(a^2+b^2)x+ab=0 \iff x^2-\frac{a^2+b^2}{ab}x+1=0 \iff x^2-(\frac{a}{b}+\frac{b}{a})x+1=0$.
Let $r_1$ and $r_2$ be the (possibly equal, possibly complex) roots of the quadratic function $x^2-(\frac{a}{b}+\frac{b}{a})x+1$. Then their product is $1$ and their sum is $\frac{a}{b}+\frac{b}{a}$. This shows that the roots are $\frac{b}{a}$ and $\frac{a}{b}$.
Therefore, $x^2-(\frac{a}{b}+\frac{b}{a})x+1=(x-\frac{a}{b})(x-\frac{b}{a})$ so
$$abx^2-(a^2+b^2)x+ab = ab(x-\frac{a}{b})(x-\frac{b}{a}) = (ax-b)(bx-a)$$
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