\(\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-4\)
\(=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)
Đặt \(x^2+6x+5=t\) thì biểu thức trở thành:
\(t\left(t+3\right)-4=\left(t-1\right)\left(t+4\right)\)
\(=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)