Matemática, perguntado por pedrofernandes45jzzs, 8 meses atrás

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Soluções para a tarefa

Respondido por marcamte
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Resposta:

f(x) = (x - 1) (x -2) (x - 1 - i) (x - 1 + i)

Explicação passo-a-passo:

f(x) = x^{4} - 5x^3 + 10x^2 - 10x + 4 funcao do 4º grau tem 4 raizes, e vc pode escreve-la como: \\\\f(x) = (x - r1) (x - r2) (x - r3) (x - r4)\\\\\\r1 = 1, r2 = 2\\\\f(x) = (ax^2 + bx + c) (x-1) (x-2)\\\\f(x) = (ax^2 + bx + c) (x^2-3x+2) = x^{4} - 5x^3 + 10x^2 - 10x + 4 \\\\ax^2 x^2 = ax^4 = x^4\\\\a = 1\\\\ax^2 (-3x) + bx(x^2) = -5x^3 -> a = 1,\\\\-3x^3 + bx^3 = -5x^3\\\\b-3 = -5\\\\b = -2\\\\ax^2(2) + bx(-3x) + cx^2 = 10x^2\\\\a = 1 e b = -2\\\\2x^2 + 6x^2 + cx^2 = 10 x^2\\\\

8 + c = 10\\\\c = 2\\

f(x) = (ax^2 + bx + c) (x-1) (x-2) = 0\\\\a = 1, b = -2 , c=2\\\\(x^2 -2x + 2) = 0\\\\d = (-2)^2 - 4(1)(2)\\\\d = 4 - 8\\\\d = -4\\\\r = \frac{-b±\sqrt{d} }{2a}\\\\r3 = \frac{2 + \sqrt{-4}}{2} \\\\r3 = \frac{2 + 2i}{2} \\\\r3 = 1 + i\\\\r4 = \frac{2 -  \sqrt{-4}}{2} \\\\r4 = \frac{2 - 2i}{2} \\\\r4 = 1 -  i\\

f(x) = (x - 1) (x -2) (x - 1 - i) (x - 1 + i)

verificando:

(x-1) (x-2) = x^2 - 3x + 2(I)\\\\\\(x - 1-i) (x - 1+i) = x^2 - x + xi - x + 1 - i - xi + i - i^2\\\\(x - 1-i) (x - 1+i) = x^2 - x - x + 1 - i^2 \\\\ (i^2 = -1)\\\\(x - 1-i) (x - 1+i) = x^2 - 2x + 1 - (-1)\\\\(x - 1-i) (x - 1+i) = x^2 - 2x + 2(II)\\\\\\(x - 1) (x -2) (x - 1 - i) (x - 1 + i) = (x^2 - 3x + 2)(x^2 - 2x + 2)\\\\(x - 1) (x -2) (x - 1 - i) (x - 1 + i) = (x^4 - 2x^3 + 2x^2 - 3x^3 + 6x^2 - 6x + 2x^2 - 4x + 4)\\\\(x - 1) (x -2) (x - 1 - i) (x - 1 + i) = x^4 - 5x^3 + 10x^2 - 10x + 4,  CQD

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