Đặt: \(x^2+10x+21=t\)
Ta có: \(A=\left(\left(x+2\right)\left(x+8\right)\right)\left(\left(x+4\right)\left(x+6\right)\right)+2008\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\)
Thay t vào ta được: \(A=\left(t-5\right)\left(t+3\right)+2008=t^2-2t+15+2008=t^2-2t+2023\)
Vậy A chia t dư 2023