\(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-\left(2x^2+4x\right)-3\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3\)
Đặt \(x^2+2x=t\) thì:
Biểu thức \(=t^2-2t-3=\left(t-3\right)\left(t+1\right)\)
\(=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x+1\right)^2\)