faktorisieren x^4-6x^3+9x^2+4x-12

Lösung

faktorisieren

x^{4} - 6 x^{3} + 9 x^{2} + 4 x - 12

(x + 1) (x - 2)^{2} (x - 3)

Schritte zur Lösung

x^{4} - 6 x^{3} + 9 x^{2} + 4 x - 12

Wende den rationalen Nullstellentest an

= (x + 1) \frac{x ^{4} - 6 x ^{3} + 9 x ^{2} + 4 x - 12}{x + 1}

\frac{x ^{4} - 6 x ^{3} + 9 x ^{2} + 4 x - 12}{x + 1} = x^{3} - 7 x^{2} + 16 x - 12

= (x + 1) (x^{3} - 7 x^{2} + 16 x - 12)

Faktorisiere

x^{3} - 7 x^{2} + 16 x - 12 : (x - 2) (x - 2) (x - 3)

= (x + 1) (x - 2) (x - 2) (x - 3)

Wende Exponentenregel an:

aa = a^{2}

(x - 2) (x - 2) = (x - 2)^{2}

= (x + 1) (x - 2)^{2} (x - 3)

5 y > 25

2.5 (10 x - 16) = 5

s im pl i f y 4 - \frac{5}{4 m + 8}

s im pl i f y \frac{\frac{1}{( x + h ) ^{2}} - \frac{1}{x ^{2}}}{h}

f a c t or 14 m^{4} - 75 m^{2} - 11