faktorisieren x^5-3x^4+12x^3-28x^2+27x-9

Lösung

faktorisieren

x^{5} - 3 x^{4} + 12 x^{3} - 28 x^{2} + 27 x - 9

(x - 1)^{3} (x^{2} + 9)

Schritte zur Lösung

x^{5} - 3 x^{4} + 12 x^{3} - 28 x^{2} + 27 x - 9

Wende den rationalen Nullstellentest an

= (x - 1) \frac{x ^{5} - 3 x ^{4} + 12 x ^{3} - 28 x ^{2} + 27 x - 9}{x - 1}

\frac{x ^{5} - 3 x ^{4} + 12 x ^{3} - 28 x ^{2} + 27 x - 9}{x - 1} = x^{4} - 2 x^{3} + 10 x^{2} - 18 x + 9

= (x - 1) (x^{4} - 2 x^{3} + 10 x^{2} - 18 x + 9)

Faktorisiere

x^{4} - 2 x^{3} + 10 x^{2} - 18 x + 9 : (x - 1) (x - 1) (x^{2} + 9)

= (x - 1) (x - 1) (x - 1) (x^{2} + 9)

Wende Exponentenregel an:

aaa = a^{3}

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

= (x - 1)^{3} (x^{2} + 9)

5 y - 6 = 29 - 2 y

f a c t or 15 b^{2} - 7 b - 2

2 x^{2} + 7 x - 9 = 0

x^{2} + 5 x \geq 0

4 y^{2} - 81 = 0