Question

Con resolución aporte 10 puntos

Answer 1

$P(x)=x^7+x^6+x^5+x^3-1\\ \\ P(x)=x^5(x^2+x+1)+(x^3-1)\\ \\ P(x)=x^5(x^2+x+1)+(x-1)(x^2+x+1)\\ \\ P(x)=(x^2+x+1)(x^5+x-1)\\ \\ P(x)=(x^2+x+1)[(x^5+x^2)-(x^2-x+1)]\\ \\ P(x)=(x^2+x+1)[x^2(x^3+1)-(x^2-x+1)]\\ \\ P(x)=(x^2+x+1)[x^2(x+1)(x^2-x+1)-(x^2-x+1)]\\ \\ P(x)=(x^2+x+1)(x^2-x+1)[x^2(x+1)-1]$


$P(x)=(x^2+x+1)(x^2-x+1)(x^3+x^2-1)\\ \\ P(x)=(x^2+x+1)(x^2-x+1)\left(x-\frac{\sqrt[3]{100-12\sqrt{69}}+\sqrt[3]{100+12\sqrt{69}}}{6}+\frac{1}{3}\right)...\\ \\ \left[x^2+\left(\frac{\sqrt[3]{100-12\sqrt{69}}+\sqrt[3]{100+12\sqrt{69}}}{6}+\frac{2}{3}\right)x+\frac{\sqrt[3]{100-12\sqrt{69}}+\sqrt[3]{100+12\sqrt{69}}}{18}\right.+ \\ \\ \\ \left. ~~~~~~~~~~+\frac{\sqrt[3]{2492-300\sqrt{69}}+\sqrt[3]{2492+300\sqrt{69}}}{18}\right]$

Ojo que $x^3+x^2-1$ no es factor primo.