Giải thích các bước giải:
a) `f(x)=x^2+ax+3`
`=(x^2-x)+ax-x+3`
`=x(x-1)+[(a-1)x-(a-1)]+a-1+3`
`=x(x-1)+(a-1)(x-1)+a+2`
`=(x+a-1)(x-1)+a+2`
Để `f(x)vdots Q(x)=>(x+a-1)(x-1)+a+2 vdots x-1`
Mà `(x+a-1)(x-1)vdots x-1=>a+2=0=>a=-2`
b) `f(x)=x^3+ax^2+2x-3a`
`=(x^3+2x^2)+ax^2-2x^2+2x-3a`
`=x^2(x+2)+[(a-2)x^2+2(a-2)x]+2x-2(a-2)x-3a`
`=x^2(x+2)+x(a-2)(x+2)+(2-2a+4)x-3a`
`=(x+2)[x^2+x(a-2)]+[(6-2a)x+2(6-2a)]-2(6-2a)-3a`
`=(x+2)[x^2+x(a-2)]+(6-2a)(x+2)-12+4a-3a`
`=(x+2)[x^2+x(a-2)+6-2a]+a-12`
Để `f(x)vdots Q(x)=>(x+2)[x^2+x(a-2)+6-2a]+a-12 vdots x+2`
Mà `(x+2)[x^2+x(a-2)+6-2a] vdots x+2=>a-12=0=>a=12`
