Lời giải:
\(2x^7+x^5+2=2(x^7-x)+x^5-x^2+(x^2+2x+2)\)
\(=2x(x^6-1)+x^2(x^3-1)+(x^2+x+1)+(x+1)\)
\(=2x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)+(x+1)\)
\(=(x^3-1)(2x^4+2x+x^2)+(x^2+x+1)+(x+1)\)
\(=(x-1)(x^2+x+1)(2x^4+2x+x^2)+(x^2+x+1)+(x+1)\)
\(=(x^2+x+1)[(x-1)(2x^4+2x+x^2)+1]+(x+1)\)
Do đó \(2x^7+x^5+2: (x^2+x+1)=(x-1)(2x^4+2x+x^2)+1\) dư $x+1$