`5a-3b+2c=0`
`P(x)=ax^2+bx+c`
`P(-1)=a.(-1)^2+b.(-1)+c=a-b+c`
`P(-2)=a.(-2)^2+b.(-2)+c=4a-2b+c`
Ta có:
`\qquad P(-1)+P(-2)=(a-b+c)+(4a-2b+c)`
`=5a-3b+2c=0`
`=>P(-2)=-P(-1)`
`=>P(-1).P(-2)`
`=P(-1).[-P(-1)]=-[P(-1)]^2`
`=-(a-b+c)^2\le 0` với mọi `a;b;c`
Vậy nếu `5a-3b+2c=0` thì:
`\qquad P(-1).P(-2)\le 0` (đpcm)