`text{C1:Đặt f(x)=}` `2x^3-5x^2+10x-m`
`text{Vì 2x-1 có nghiệm là} ` `1/2` `text{nên theo định lí Bozu thì dư là:}`
`f(1/2)=2.(1/2)^3-5.(1/2)^2+10. 1/2-m`
`=2. 1/8-5. 1/4+5-m`
`=4-m`
`⇒f(x) \vdots (2x-1)⇔4-m=0`
`⇔m=4.`
`text{C2:Ta có:}`
`2x^3-5x^2+10x-m`
`=2x^3-x^2-4x^2+2x+8x-4+4-m`
`=(2x^3-x^2)-(4x^2-2x)+(8x-4)+(4-m)`
`=x^2(2x-1)-2x(2x-1)+4(2x-1)+(4-m)`
`=(x^2-2x+4)(2x-1)+(4-m)`
`⇒(2x^3-5x^2+10x-m) \vdots (2x-1)`
`⇔4-m=0`
`⇔m=4.`
