Đặt $a=x^2+7x+9$
Ta có:
`\qquad (x+2)(x+3)(x+4)(x+5)-24`
`=(x+2)(x+5)(x+3)(x+4)-24`
`=(x^2+5x+2x+10)(x^2+4x+3x+12)-24`
`=(x^2+7x+10)(x^2+7x+12)-24`
`=(x^2+7x+9+1)(x^2+7x+9+3)-24`
`=(a+1)(a+3)-24`
`=a^2+3a+a+3-24`
`=a^2+4a-21`
`=a(a+4)-21`
Vì $a(a+4)$ chia hết cho $a$
`\qquad -21` chia $a$ dư $(-21)$
`=>a(a+4)-21` chia $a$ dư $(-21)$
Vậy $(x+2)(x+3)(x+4)(x+5)-24$ chia cho $(x^2+7x+9)$ dư $(-21)$