• Matéria: Matemática
  • Autor: biancapaulina2019
  • Perguntado 5 anos atrás

RESOLVA AS DIVISÕES A SEGUIR PELO TEOREMA DO RESTO:

A) P(X) =X 4 + 3X 3 – 4X 2 + 8X – 12 DIVIDIDO POR D(X) = X – 2

B) P(X) =X 4 + 3X 3 – 4X 2 + 8X – 8 DIVIDIDO POR D(X) = X – 1

C) P(X) =X 4 + 6X 3 – 4X 2 + 8X – 6 DIVIDIDO POR D(X) = X – 3

D) P(X) =X 4 + 2X 3 – 4X 2 + 8X – 10 DIVIDIDO POR D(X) = X – 1

E) P(X) =X 4 + 7X 3 – 10X 2 + 15X – 10 DIVIDIDO POR D(X) = X – 5

F) P(X) =X 4 + 2X 3 – 2X 2 + 8X – 6 DIVIDIDO POR D(X) = X – 2

Respostas

respondido por: Menelaus
1

A)

x-2 ≡ 0(mod x-2)

x ≡ 2(mod x-2)

x^2 ≡ 4(mod x-2)

x^3 ≡ 8(mod x-2)

x^4 ≡ 16(mod x-2)

Resto: 16 + 24 + 16 + 16 - 12 = 48 + 12 = 60

B)

x-1 ≡ 0(mod x-1)

x ≡ 1(mod x-1)

x^2 ≡ 1(mod x-1)

x^3 ≡ 1(mod x-1)

x^4 ≡ 1(mod x-1)

Resto: 1 + 3 - 4 + 8 - 8 = 0

C)

x-3 ≡ 0(mod x-3)

x ≡ 3(mod x-3)

x^2 ≡ 9(mod x-3)

x^3 ≡ 27(mod x-3)

x^4 ≡ 81(mod x-3)

Resto: 81 + 162 - 36 + 24 - 6 = 225

D)

x-1 ≡ 0(mod x-1)

x ≡ 1(mod x-1)

x^2 ≡ 1(mod x-1)

x^3 ≡ 1(mod x-1)

x^4 ≡ 1(mod x-1)

Resto: 1 + 2 - 4 + 8 - 10 = -3

E)

x-5 ≡ 0(mod x-5)

x ≡ 5(mod x-5)

x^2 ≡ 25(mod x-5)

x^3 ≡ 125(mod x-5)

x^4 ≡ 625(mod x-5)

Resto: 625 + 875 - 100 + 40 - 10 = 1430

F)

x-2 ≡ 0(mod x-2)

x ≡ 2(mod x-2)

x^2 ≡ 4(mod x-2)

x^3 ≡ 8(mod x-2)

x^4 ≡ 16(mod x-2)

Resto: 16 + 16 - 8 + 8 - 6 = 26

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