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)?
 A: Using the quadratic formula is a good approach, and everything you have done so far is correct. The modulus/absolute value can be replaced by normal brackets as there is a $\pm$ preceding it, and the numbers involved here are real. Notice that:
$$\frac {a^2 +b^2 + (a^2 - b^2)}{2ab} = \frac ab, \quad \frac {a^2 +b^2 - (a^2 - b^2)}{2ab} = \frac ba$$
hence the roots are $\dfrac ab$ and $\dfrac ba$, and we are looking at the factors $(bx- a)$ and $(ax - b)$. Now:
$$(bx-a)(ax- b) = abx^2 - a^2 x - b^2 x + ab$$
which is exactly what we are trying to factorize.
A: 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.
A: 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
$$abx^2-(a^2+b^2)x+ab = ab(x-\frac{a}{b})(x-\frac{b}{a}) = (ax-b)(bx-a)$$
A: Everything is OK.
First: Accept
$$a^2≥b^2$$
Then accept
$$a^2≤b^2$$
You will see that the roots have not changed.
We have
$$abx^2-(a^2+b^2)x+ab=(a x - b) (b x - a)$$

Small Supplement:
Note that,
$$\begin{cases} ±|a^2-b^2|=±(a^2-b^2), a^2≥b^2 \\ ±|a^2-b^2|=±(b^2-a^2), a^2≤b^2 \end{cases}$$
$$\implies \pm(a^2-b^2)=\mp(b^2-a^2)$$
Thus,
$$±|a^2-b^2|=±(a^2-b^2).$$
A: 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.
