Find coefficient of $x^2$ in a complicated expansion 
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$

I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there I am stuck when it comes to expanding $(1+2x+3x^2)^6$ as I can't see an easy way to use the binomial theorem (for example by factorising $1+2x+3x^2$, which I can't do as there are no real roots). Is there a way to find the coefficient of $x^2$ in this expansion (specifically $(3x^2+2x+6)^6$ as I can get the answer from there)?

Let $A=\big[(1+2x+3x^2)^6+(1+4x^3)^5\big]$.
$$\begin{align}
\color{green}{[x^2]}A
&= \color{green}{[x^2]}(1+2x+3x^2)^6+\color{green}{[x^2]}(\underbrace{1+4x^3}_{\color{blue}{\text{no $x,x^2$ terms}}})^5\\
&=\color{green}{[x^2]}\bigg[\binom 6{5,0,1}1^5(2x)^0(3x^2)^1+\binom 6{4,2,0}1^4(2x)^2(3x^2)^0\bigg]+\color{blue}0\\
&=\frac{6!}{5!\;0!\;1!}\cdot 1^52^03^1+\frac {6!}{4!\;2!\;0!}\cdot 1^42^23^0\\
&=78\\\\
\color{green}{[x^0]}A
&=\color{green}{[x^0]}(1+2x+3x^2)^6+\color{green}{[x^0]}(1+4x^3)^5\\
&=1+1\\
&=2\\\\
\color{green}{[x^2]}(4-x^2)A
&=4\cdot \big(\color{green}{[x^2]}A\big)-\big(\color{green}{[x^0]}A\big)\\
&=4\cdot 78-2\\
&=\color{red}{310}
\end{align}$$
See wiki entry on the Multinomial Theorem for more details.
