<s>[INST] What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.
I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.
if I'm correct here then the answer should be $(16*270)+(24*243)$? but this does not appear to be the case.
any assistance appreciated.
 [/INST] Using the binomial theorem we have:
$$(x+2)^4(x+3)^5 = \left( \sum_{k=0}^{4}\binom{4}{k}x^k2^{4-k}\right) \left(\sum_{j=0}^{5}\binom{5}{j}x^j3^{5-j} \right).$$
We get $x^2$ by taking an $x$ term from each sum or an $x^2$ and $x^0$ term from one or the other sum. So the coefficient of $x^2$ is $$\overbrace{\binom{4}{1}2^3\binom{5}{1}3^4}^{x\cdot x} + \overbrace{2^4\binom{5}{2}3^3}^{x^0\cdot x^2} + \overbrace{\binom{4}{2}2^23^5}^{x^2\cdot x^0}$$
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