Ta có: \(A=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+2028\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+2028\)
Đặt: \(x^2+8x+12=t\) ta có: \(x^2+8x+7=t-5\) và \(x^2+8x+15=t+3\)
Ta có: \(A=\left(t+3\right)\left(t-5\right)+2028=t^2-2t+2013\)chia t dư 2013
Vậy A chia x2 + 8x + 12 dư 2013