faktorisieren a^4-b^4+2b^2-1

Lösung

faktorisieren

a^{4} - b^{4} + 2 b^{2} - 1

(a^{2} + b^{2} - 1) (a^{2} - b^{2} + 1)

Schritte zur Lösung

a^{4} - b^{4} + 2 b^{2} - 1

Faktorisiere

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

= a^{4} - (b + 1)^{2} (b - 1)^{2}

a^{4} = (a^{2})^{2}

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

= (a^{2})^{2} - ((b + 1) (b - 1))^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

(a^{2})^{2} - ((b + 1) (b - 1))^{2} = (a^{2} + (b + 1) (b - 1)) (a^{2} - (b + 1) (b - 1))

= (a^{2} + (b + 1) (b - 1)) (a^{2} - (b + 1) (b - 1))

Vereinfache

(a^{2} + (b + 1) (b - 1)) (a^{2} - (b + 1) (b - 1)) : (a^{2} + b^{2} - 1) (a^{2} - b^{2} + 1)

= (a^{2} + b^{2} - 1) (a^{2} - b^{2} + 1)

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

e x p an d (x - y)^{2} - 9

s im pl i f y \frac{2 sin ( x ) - 4 sin ( x ) - cos ( x )}{2}

s im pl i f y (8 x^{2} + 7 x + 10) + (- 2 x^{3} + 7 x + 4)

s im pl i f y l^{- 1} \cdot ⎩ ⎨ ⎧ \frac{24}{( ( s - 1 ) ^{5} )} ⎭ ⎬ ⎫