faktorisieren x^{10}+x^8+x^6+x^4+x^2+1

Lösung

faktorisieren

x^{10} + x^{8} + x^{6} + x^{4} + x^{2} + 1

(x^{2} + x + 1) (x^{2} - x + 1) (x^{2} + 1) (x^{4} - x^{2} + 1)

Schritte zur Lösung

x^{10} + x^{8} + x^{6} + x^{4} + x^{2} + 1

Angenommen

u = x^{2}

= u^{5} + u^{4} + u^{3} + u^{2} + u + 1

Faktorisiere

u^{5} + u^{4} + u^{3} + u^{2} + u + 1 : (u^{2} + u + 1) (u + 1) (u^{2} - u + 1)

= (u^{2} + u + 1) (u + 1) (u^{2} - u + 1)

Setze in

u = x^{2}

ein

= (x^{4} + x^{2} + 1) (x^{2} + 1) (x^{4} - x^{2} + 1)

Faktorisiere

x^{4} + x^{2} + 1 : (x^{2} + 1 + x) (x^{2} + 1 - x)

= (x^{2} + 1 + x) (x^{2} + 1 - x) (x^{2} + 1) (x^{4} - x^{2} + 1)

= (x^{2} + x + 1) (x^{2} - x + 1) (x^{2} + 1) (x^{4} - x^{2} + 1)

s im pl i f y \frac{2}{x + 3} - \frac{1}{x ^{2} + 7 x + 12}

\frac{3}{4} \div \frac{5}{7}

s im pl i f y (sec^{2} (x) + csc^{2} (x)) - (tan^{2} (x) + cot^{2} (x))

s im pl i f y - 2 3 64 + (- 5^{2})

e x p an d (6 x - x^{2})^{2}