\(A=\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
Đặt \(t=x^2+3x+1\) thì A thành
\(t\left(t-4\right)-5=t^2-4t-5\)
\(t^2-5t+t-5=t\left(t-5\right)+\left(t-5\right)\)
\(=\left(t-5\right)\left(t+1\right)=\left(x^2+3x+1-5\right)\left(x^2+3x+1+1\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
đặt a=x^2+3x+1
phương trình đã cho thành phương trình: a(a-4)-5
=a^2-4a-5
=a^2+a-5a-5
= a(a+1)-5(a+1)
=(a-5)(a+1)
=(x^2+3x-4)(x^2+3x+2)
=(x-1)(x+1)(x+2)(x+4)