Ta có:

`f(x) = ax^2 + bx + c`

`⇒ f(-2) = 4a − 2b + c`

`⇒ f(3) = 9a + 3b + c`

Suy ra:

`f(-2) + f(3) = (4a − 2b + c) + (9a + 3b + c) `

`= 13a + b + 2c `

Theo giả thiết: `13a + b + 2c = 0 `

`⇒ f(-2) + f(3) = 0`

`⇒ f(-2) = −f(3)`

Suy ra

`f(-2) . f(3) = −[f(3)]^2`

Vì `[f(3)]^2 ≥ 0`

`⇒ f(-2) . f(3) ≤ 0 (đpcm)`

Vậy `f(-2) . f(3) ≤ 0`