Đặt $x^{2}$ + $4x + 8 = t^{}$
Ta có: $t^{2}$ $+ 3xt + 2x^{2}$
$= t^{2}$ $+ 2xt + xt + 2x^{2}$
$ = t ( t + 2x ) + x ( t + 2x )^{}$
$( t + x) ( t + 2x)^{}$
Vậy $(x^{2}$ $ + 4x + 8 )^{2}$ $+ 3x ( x^{2}$ $+ 4x + 8) + 2x^{2}$ $= ( x^{2}$ $ + 4x + 8 +x)^{}$ $(x^{2}$ + $4x + 8 + 2x)^{}$
$= ( x^{2}$ $+ 5x + 8 ) ( x^{2}$ $+ 6x + 8)^{}$ $= (x^{2}$ $+ 5x + 8 ) ( x + 2 ) ( x + 4 )^{}$
XIN CTLHN NHA BẠN^ ^