\(B=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=t\)
Ta được:
\(B=t\left(t+1\right)-6\)
\(B=t^2+t-6\)
\(B=t^2+3t-2t-6\)
\(B=t\left(t+3\right)-2\left(t+3\right)\)
\(B=\left(t+3\right)\left(t-2\right)\)
\(B=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Vậy......