簡約 (x+4)(x+1-3i)^2

解

簡約

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

(x^{3} + 6 x^{2} - 32) + (- 24 - 30 x - 6 x^{2}) i

解答ステップ

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

= (x + 1 - 3 i)^{2} (x + 4)

分配法則を適用する:

a (b + c) = ab + a c

(x + 1 - 3 i)^{2} (x + 4) = (x + 1 - 3 i)^{2} x + (x + 1 - 3 i)^{2} \cdot 4

= (x + 1 - 3 i)^{2} x + (x + 1 - 3 i)^{2} \cdot 4

簡素化

(x + 1 - 3 i)^{2} x + (x + 1 - 3 i)^{2} \cdot 4 : - 24 i + x^{3} + 6 x^{2} - 30 x i - 6 x^{2} i - 32

= - 24 i + x^{3} + 6 x^{2} - 30 x i - 6 x^{2} i - 32

標準的な複素数形式で書き直す:

(x^{3} + 6 x^{2} - 32) + (- 24 - 30 x - 6 x^{2}) i

= (x^{3} + 6 x^{2} - 32) + (- 24 - 30 x - 6 x^{2}) i