Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$ 
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what  should I do to factorize this polynomial ?  
(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)  

$(x+1)(x+2)(x+3)(x+6)-3x^2$
$\rightarrow(x+1)(x+6)(x+2)(x+3)-3x^2$
$\rightarrow(x^2+x+6x+6)(x^2+2x+3x+6)-3x^2$
$\rightarrow(x^2+7x+6)(x^2+5x+6)-3x^2$
$\rightarrow(x^2+6+7x)(x^2+6+5x)-3x^2$
Put :- $x^2+6=y$ eq($1$) . So :-
$(y+7x)(y+5x)-3x^2$
$\rightarrow y^2+7xy+5xy+35x^2-3x^2$
$\rightarrow y^2+12xy+32x^2$
$\rightarrow y^2+4xy+8xy+32x^2$
$\rightarrow y(y+4x)+8x(y+4x)$
$\rightarrow (y+4x)(y+8x)$
Put value of $y$ from eq(1), we get :-
$\rightarrow (x^2+6+4x)(x^2+6+8x)$
