展開する 2((x+1)(x+2)(x-1)(x-1)^2)

解

展開する

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

2 x^{5} - 8 x^{3} + 4 x^{2} + 6 x - 4

解答ステップ

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

括弧を削除する:

(a) = a

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

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

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

数を足す:

1 + 2 = 3

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

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

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

拡張

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

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

拡張

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

= 2 (x^{5} - 4 x^{3} + 2 x^{2} + 3 x - 2)

拡張

2 (x^{5} - 4 x^{3} + 2 x^{2} + 3 x - 2) : 2 x^{5} - 8 x^{3} + 4 x^{2} + 6 x - 4

= 2 x^{5} - 8 x^{3} + 4 x^{2} + 6 x - 4