Question

quem consegue resolver ??

Answer 1

$N = \frac{(x^2-2x-15)(x^3+5x^2+25x)}{337(x^4+3x^3-125x-375)} \\ \\ N= \frac{x^5+5x^4+25x^3-2x^4-10x^3-50x^2-15x^3-75x^2-375x}{337(x^4+3x^3-125x-375)} \\ \\ N= \frac{x^5+3x^4-125x^2-375x}{337(x^4+3x^3-125x-375)} \\ \\ N= \frac{x(x^4+3x^3-125x-375)}{337(x^4+3x^3-125x-375)} \\ \\ N= \frac{x}{337}$

Substituindo x= 1685, temos:

$N = \frac{x}{337} = \frac{1685}{337} =5$

Sen N=5 é um multiplo de 5. letra B)