\(a,\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (sửa \(2x\rightarrow2x^2\)
Đặt \(x^2+4x+8=a\)
\(=a^2+3ax+2x=a^2+ax+2ax+2x^2=\left(a+x\right)\left(a+2x\right)\\ =\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x\)
\(=\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)
\(=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)