Ta có:
`\qquad 2x^3-x^2+2x+a`
`=2x^3-6x^2+5x^2-15x+17x-51+a+51`
`=2x^2(x-3)+5x(x-3)+17(x-3)+a+51`
`=(x-3)(2x^2+5x+17)+a+51`
Để `2x^3-x^2+2x+a \vdots x-3`
`<=> (x-3)(2x^2+5x+17)+a+51 \vdots x-3`
Mà `(x-3)(2x^2+5x+17) \vdots x-3`
`=> a+51 \vdots x-3`
`<=> a+51=0 `
`<=> a=-51`
Vậy `a=-51`