Find coefficient of $x^2$ in a complicated expansion

Question

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)?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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 $310$

Answer 5

There are a number of different approaches here.

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

2. 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

3. 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.

Answer 6

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.