簡約 B(x+1)(x+i)(x-i)^2+D(x+1)(x-i)(x+i)^2

解

簡約

B (x + 1) (x + i) (x - i)^{2} + D (x + 1) (x - i) (x + i)^{2}

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

解答ステップ

B (x + 1) (x + i) (x - i)^{2} + D (x + 1) (x - i) (x + i)^{2}

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

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

= B (x + 1) (x + i) (x^{2} - 2 x i - 1) + D (x + 1) (x - i) (x^{2} + 2 x i - 1)

拡張

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

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

拡張

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

= - B i + B x^{4} - B x^{3} i + B x^{2} + B x^{3} - B x i - B x^{2} i + B x + D i + x^{4} D + x^{3} D i + x^{2} D + x^{3} D + D x i + x^{2} D i + D x

標準的な複素数形式で書き直す:

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

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