Find the Coefficient of $x^3$ in $(2+x^2)^3 (3+2x)^7$

Question

I'm asked to find the coefficient of $x^3$ in $(2+x^2)^3 (3+2x)^7$.

For a simple problem like finding $x^2y^3$ in $(x+y)^5$ I can solve easily using binomial theorem.

But I have no idea how to go about approaching this one using binomial theorem, since I'm given two binomials(Each raised to separate exponents), multiplied by each other

Any help would be appreciated.

Answer 1

This is about to get really REALLY tedious.

$ \sum\limits_{k=0}^{3} \binom{3}{k}(2)^{3-k}(x^2)^k \times \sum\limits_{m=0}^{7} \binom{7}{m}(3)^{7-m}(2x)^m $

Rearrange:

$ \sum\limits_{k=0}^{3}\sum\limits_{m=0}^{7} \binom{3}{k}\binom{7}{m} (2)^{3-k + m} (3)^{7-m}(x)^{2k+m} $

Now you have just do this:

When 2k + m = 3

Case 1: k = 1, m = 1

Case 2: k = 0, m = 3

You plug that into your formula (remove the sigmas) and add up all the cases, so:

$ \binom{3}{1}\binom{7}{1} (2)^{3- 1 + 1} (3)^{7-1}(x)^{2+ 1} + \binom{3}{0}\binom{7}{3} (2)^{3- 0 + 3} (3)^{7-3}(x)^{2(0)+ 3} $

Good luck have fun :)