Binomische Formeln

Die binomischen Formeln

Theorem 1: Die binomischen Formeln

Für a,b∈Ra,b \in \R und n∈N0n \in \N_0 gelten die drei sogenannten binomischen Formeln:

(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2(a+b)(a−b)=a2−b2\begin{align} (a+b)^2 &= a^2+2ab+b^2\\ (a-b)^2 &= a^2-2ab+b^2\\ (a+b)(a-b) &= a^2-b^2 \end{align}
Proof

Durch Ausmultiplizieren erhält man obige Identitäten.

Exercise 1: Multiplizieren oder faktorisieren

Multipliziere aus oder faktorisiere.

a) (x+y)2(x+y)^2

b) (x−2y)2(x-2y)^2

c) (2x−y)2(2x-y)^2

d) (−x3+y)2(-x^3+y)^2

e) (2x−y3)(2x+y3)(2x-y^3)(2x+y^3)

Solution

a) (x+y)2=x2+2xy+y2(x+y)^2 = x^2+2xy+y^2

b) (x−2y)2=x2−4xy+4y2(x-2y)^2 = x^2-4xy+4y^2

c) (2x−y)2=4x2−4xy+y2(2x-y)^2 = 4x^2-4xy+y^2

d) (−x3+y)2=(y−x3)2=y2−2x3y+x6(-x^3+y)^2 = (y-x^3)^2 = y^2-2x^3y+x^6

e) (2x−y3)(2x+y3)=4x2−y6(2x-y^3)(2x+y^3) = 4x^2-y^6

Exercise 2: Dritte binomische Formel

Berechne:

a) (78z2+1)(78z2−1)(\tfrac{7}{8} z^2 + 1)(\tfrac{7}{8} z^2 - 1)

b) (−m3+m)(m3+m)(-m^3 + m)(m^3 + m)

c) (−xy−13)(−xy+13)(-xy - 13)(-xy + 13)

d) (1.4i−1.7)(1.4i+1.7)(1.4i - 1.7)(1.4i + 1.7)

Faktorisiere:

e) x2−4x^2-4

f) −x2+1-x^2+1

g) 4u6−v44u^6-v^4

Solution

a) 4964z4−1\tfrac{49}{64} z^4 - 1

b) −m6+m2-m^6 + m^2

c) x2y2−169x^2y^2 - 169

d) 1.96i2−2.891.96i^2 - 2.89

e) x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2)

f) −x2+1=1−x2=(1−x)(1+x)-x^2+1 = 1-x^2 = (1-x)(1+x)

g) 4u6−v4=(2u3)2−(v2)2=(2u3−v2)(2u3+v2)4u^6-v^4 = (2u^3)^2-(v^2)^2 = (2u^3-v^2)(2u^3+v^2)

Exercise 3: Dritte Potenzen

Multipliziere aus.

a) (a−b)3(a - b)^3

b) (4g−1)3(4g - 1)^3

c) (m+16)3(m + \tfrac{1}{6})^3

d) (x2−y2)3(x^2 - y^2)^3

Solution

a) (a−b)3=(a−b)2(a−b)=(a2−2ab+b2)(a−b)=a3−3a2b+3ab2−b3(a - b)^3 = (a-b)^2(a-b) = (a^2-2ab+b^2)(a-b) = a^3 - 3a^2b + 3ab^2 - b^3

b) (4g−1)3=64g3−48g2+12g−1(4g-1)^3 = 64g^3 - 48g^2 + 12g - 1

c) (m+16)3=m3+12m2+112m+1216(m + \tfrac{1}{6})^3 = m^3 + \tfrac{1}{2} m^2 + \tfrac{1}{12} m + \tfrac{1}{216}

d) (x2−y2)3=x6−3x4y2+3x2y4−y6(x^2 - y^2)^3 = x^6 - 3x^4y^2 + 3x^2y^4 - y^6