Question

hola necesito ayuda en esto tema es d factorizacion_aiuda pls les doy alternativas_ah y con explicacion v'''''''':​

Answer 1

Respuesta:

b)

Explicación paso a paso:

$x^6-7x^3-1$

$\mathrm{Utilizar\:el\:teorema\:de\:la\:raiz\:racional}\\a_0=1,\:\quad a_n=8\\\mathrm{Los\:divisores\:de\:}a_0:\quad 1,\:\quad \mathrm{Los\:divisores\:de\:}a_n:\quad 1,\:2,\:4,\:8\\\mathrm{Por\:lo\:tanto,\:verificar\:los\:siguientes\:numeros\:racionales:\quad }\pm \frac{1}{1,\:2,\:4,\:8}\\\frac{1}{1}\mathrm{\:es\:la\:raiz\:de\:la\:expresion,\:por\:lo\:tanto,\:factorizar\:}x-1\\=\left(x-1\right)\frac{8x^6-7x^3-1}{x-1}$

$\left(x-1\right)\frac{8x^6-7x^3-1}{x-1}\\\\=\left(x-1\right)8x^5+\frac{8x^5-7x^3-1}{x-1}\\\\=\left(x-1\right)8x^5+8x^4+\frac{8x^4-7x^3-1}{x-1}\\\\=\left(x-1\right)8x^5+8x^4+8x^3+\frac{x^3-1}{x-1}\\\\=\left(x-1\right)8x^5+8x^4+8x^3+x^2+\frac{x^2-1}{x-1}\\\\=\left(x-1\right)8x^5+8x^4+8x^3+x^2+x+\frac{x-1}{x-1}\\\\=\left(x-1\right)8x^5+8x^4+8x^3+x^2+x+1\\\\=\left(x-1\right)\left(8x^5+8x^4+8x^3\right)+\left(x^2+x+1\right)\\\\=\left(x-1\right)\left(x^2+x+1\right)+8x^3\left(x^2+x+1\right)$

$=\left(x-1\right)\left(x^2+x+1\right)\left(8x^3+1\right)\\=\left(x-1\right)\left(x^2+x+1\right)\left(2x+1\right)\left(2^2x^2-2x+1\right)\\\\=\left(x-1\right)\left(2x+1\right)\left(x^2+x+1\right)\left(4x^2-2x+1\right)$