`\qquad x^3+2x^2+ax+b`
`=(x^3+x^2+x)+(x^2+x+1)+(a-2)x+b-1`
`=x(x^2+x+1)+(x^2+x+1)+(a-2)x+b-1`
`=(x^2+x+1)(x+1)+(a-2)x+b-1`
Để `x^3+2x^2+ax+b \vdots x^2+x+1`
`=> (a-2)x+b-1 \vdots x^2+x+1`
`<=> {(a-2=0),(b-1=0):}<=>{(a=2),(b=1):}`
Vậy `(a;b)=(2;1)` thì `f(x) \vdots g(x)`