faktorisieren 2x^5-10x^4+7x^3+13x^2+3x+9

Lösung

faktorisieren

2 x^{5} - 10 x^{4} + 7 x^{3} + 13 x^{2} + 3 x + 9

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

Schritte zur Lösung

2 x^{5} - 10 x^{4} + 7 x^{3} + 13 x^{2} + 3 x + 9

Wende den rationalen Nullstellentest an

= (x + 1) \frac{2 x ^{5} - 10 x ^{4} + 7 x ^{3} + 13 x ^{2} + 3 x + 9}{x + 1}

\frac{2 x ^{5} - 10 x ^{4} + 7 x ^{3} + 13 x ^{2} + 3 x + 9}{x + 1} = 2 x^{4} - 12 x^{3} + 19 x^{2} - 6 x + 9

= (x + 1) (2 x^{4} - 12 x^{3} + 19 x^{2} - 6 x + 9)

Faktorisiere

2 x^{4} - 12 x^{3} + 19 x^{2} - 6 x + 9 : (x - 3) (x - 3) (2 x^{2} + 1)

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

Wende Exponentenregel an:

aa = a^{2}

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

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

s im pl i f y \frac{2}{7 a} + \frac{3}{5 a}

s im pl i f y \frac{25 !}{5 ! ( 25 - 5 ) !}

10 - 3 x + 10 x - 7 = 5 x - 5 + 2 x + 8

- 4 n + 11 < - 9

\frac{x - 1}{2} = 4