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)?
 A: Yeah, you have the right ideas :) $~$ This answer is the same as gt69's but packaged slightly differently:
Writing it out longform,
$$(1+2x+3x^2)^6 = (1+2x+3x^2)(1+2x+3x^2)\cdots(1+2x+3x^2).$$
There are only two ways to obtain an $x^2$ term:


*

*Pick the $2x$ from two of the factors, and $1$ for the remaining four.

*Pick the $3x^2$ from one of the factors, and $1$ for the remaining five.


The reason that binomial coefficients show up is that, for instance, $\binom 6 2$ tells you how many different ways you can pick two of the six factors.
So the $x^2$-term on the square-bracket part of the expression is $\binom 6 2 4x^2 + \binom 6 1 3x^2$, and it seems like from there you're on track for solving the rest of the problem.
A: 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.
A: Apply the Binomial Theorem twice:
$$
(a+b+c)^n = \sum_{k=0}^n \binom{n}{k} a^k (b+c)^{n-k}
$$
and now again on $(b+c)^{n-k}$.
Another way is to use multinomial theorem straight up
A: There are a number of different approaches here. 


*

*Completely work out the expansion and look for the correct term.
This is of course a lot of work.

*Since you know the that the expanded result is a polynomial in $x$,
let us say $$ P(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots $$ you
can calculate the second derivative, which gives $$ P(x) = 2 a_2 + 6
a_3 x + \dots $$ insert $x=0$, and you get twice the value of the
number you want to compute

*A minimal effort approach, would be to realise that in order to get
a term with $x^2$, you know that $4$ in the first factor needs to be combined with a quadratic term in the second factor $[\dots]$ or $-x^2$ of the first factor is combined with a constant from the second factor. If you use the same idea for finding the correct terms from the second factor you also find the correct result.
Give it a try, and if you run into problems let me know.
A: For a question like this cheating is legal :)
$\left(4-x^2\right) \left[\left(4 x^3+1\right)^5+\left(3 x^2+2 x+1\right)^6\right]=-1024 x^{17}+4096 x^{15}-2009 x^{14}-2916 x^{13}+1718 x^{12}+1844 x^{11}+15417 x^{10}+31144 x^9+33976 x^8+29832 x^7+20681 x^6+10488 x^5+4302 x^4+1428 x^3+310 x^2+48 x+8$
The answer is $310$
A: Note that$(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$ is equal to
$$4[(1+2x)^6+6\cdot 3x^2+1+o(x^2)]-x^2\cdot (1+1+o(x))\\
=4[(1+6\cdot 2x+\binom{6}{2}(2x)^2+18x^2+1+o(x^2)]-2x^2+o(x^2)\\
=8+48x+310x^2+o(x^2)$$
where we used the Binomial Theorem and the little-o notation.
