Frage zu einer Vereinfachung eines Bruchs

Question

$$\frac { (x^2-5x+6)(x^2+5x+4)}{ x^2-6x+8 }=\frac { (x-3)(x-2)(x+1)(x+4) }{ (x-2)(x-4) }=\frac { (x-3)(x+1)(x+4) }{ (x-4)}=\frac { x^3+2x^2-11x-12 }{ x-4 }=x^2+6x+\frac { 40 }{ x-4 }+13$$  Ich verstehe den letzten Schritt der Vereinfachung nicht, wie komme ich auf x^2+6x+40/(x-4)+13

Answer 1

Wir haben folgendes: $$\frac{x^3+2x^2-11x-12}{x-4}=\frac{x^3+2x^2-4x^2+4x^2-11x-12}{x-4} \\ =\frac{x^3-4x^2+6x^2-11x-12}{x-4} \\ =\frac{x^3-4x^2}{x-4}+\frac{6x^2-11x-12}{x-4} \\ =\frac{x^2(x-4)}{x-4}+\frac{6x^2-11x-12}{x-4} \\ =x^2+\frac{6x^2-11x-12}{x-4} \\ =x^2+\frac{6x^2-6\cdot 4x+6\cdot 4x-11x-12}{x-4} \\ =x^2+\frac{6x^2-24x+24x-11x-12}{x-4} \\ =x^2+\frac{6x^2-24x}{x-4}+\frac{13x-12}{x-4} \\ =x^2+6x\cdot \frac{x-4}{x-4}+\frac{13x-12}{x-4} \\ =x^2+6x+\frac{13x-12}{x-4} \\ =x^2+6x+\frac{13x-4\cdot 13+4\cdot 13-12}{x-4} \\ =x^2+6x+\frac{13x-4\cdot 13}{x-4}+\frac{4\cdot 13-12}{x-4} \\ =x^2+6x+13\frac{x-4}{x-4}+\frac{40}{x-4} \\ =x^2+6x+13+\frac{40}{x-4}$$