Since we know one root (and the coefficient of x^3 is 1), it is p(x) = x^3 - A+B+C x^2 + A*B + A*C + B*C x - A*B*C = (x-C) \cdot \left( x^2 + {\color{orange}b} x + {\color{teal}c}\right).

These coefficients {\color{orange}b} and {\color{teal}c} are given by polynomial division.

It is x^3 - A+B+C x^2 - \left ( x^2 (x- C) \right ) = -A-B x^2 , and we need to choose {\color{orange}b} = -A-B.

With this {\color{orange}b} we multiply {\color{orange}bx} and x - C and get - A-B x^2 + A*C+B*C x .

Now we subtract -A-B x^2 + A*B + A*C + B*C x - \left ( -A-B x^2 + A*C + B*C x \right ) = B*A x, and we need to choose {\color{teal}c} = A*B.

The calculation indeed works because we compute {\color{teal}c} \cdot \left( x - C \right ) = C x - C*B*A und C x + A*C*B - \left( C x + A*C*B \right) =0.

The solutions of the quadratic equation q(x) = x^2 +-A-B x +A*B = 0 we then find directly using Vieta's formulas:

{\color{blue}x_1} + {\color{red}x_2} = {\color{orange}A+B} und {\color{blue}x_1} \cdot {\color{red}x_2} = {\color{teal}A*B}

and thus {\color{blue}x_1} = A < {\color{red}x_2} = B.