CLASE 9. Resuelve: x 3 + 2x 2 + 2x + 1 = 0 Divisores de 1: 1 y – 1 1 2 2 1 1 1 1 0 x 3 +2x 2 +2x+1=(x+1)(x 2 +x+1).

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    02-Feb-2016

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  • CLASE 9

  • Resuelve:x3 + 2x2 + 2x + 1 = 0Divisores de 1: 1 y 1-1 1 2 2 1 1 -1 1 -1 1 -1 0 x3+2x2+2x+1=(x+1)(x2+x+1).

  • Veamos:x3+2x2+2x+1=(x+1)(x2+x+1)x2+x+1 no tiene descomposicin en .D = b2 4aca = 1b = 1c = 1D = 12 4.1.1D = 3.

  • x3+2x2+2x+1=(x+1)(x2+x+1)D = 3;

  • x3+2x2+2x+1=(x+1)(x2+x+1)Como

  • Por tanto:X = 1

  • Teorema fundamental del lgebra

  • x4 + 2x2 +1 = 0(x2 +1)2 = 0x2 +1 = 0x2 = 1 x1,2 = i x3,4 = i

  • a) x3 2x2 9 = 0b) y2 2y 5 = 0ESTUDIO INDIVIDUALx3 = 3x1 = 1 + 2ix2 = 1 2i

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