Ta có:
`P=x^3+x^2-11x+m`
`P=x^3-2x^2+3x^2-6x-5x+10+m-10`
`P=x^2(x-2)+3x(x-2)-5(x-2)+m-10`
`P=(x-2)(x^2+3x-5)+m-10`
Để `P \vdots Q`
`=> (x-2)(x^2+3x-5)+m-10 \vdots x-2`
Mà `(x-2)(x^2+3x-5) \vdots x-2`
`=> m-10 \vdots x-2`
`=> m-10=0`
`<=> m=10`
Vậy `m=10` thì `P \vdots Q`