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.
 A: Hint: 
$$\color{blue}{(x+2)^4} \cdot \color{green}{(x+3)^5} = \color{blue}{(x^4+\square x^3 + \square x^2+ \square x + 16)}\cdot (\color{green}{x^5+\square x^4+\square x^3+\square x^2+\square x + 243)}$$
The term for $x^2$ in the resulting final expansion could have been made by combining a constant blue term with an $x^2$ green term, an $x$ blue term with an $x$ green term, or an $x^2$ blue term with a constant green term.
A: 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 $$\overbrace{\binom{4}{1}2^3\binom{5}{1}3^4}^{x\cdot x} + \overbrace{2^4\binom{5}{2}3^3}^{x^0\cdot x^2} + \overbrace{\binom{4}{2}2^23^5}^{x^2\cdot x^0}$$
A: \begin{array}{r}
   (x+2)^4 = && \star x^4 &+  \star x^3 &+ 24x^2 &+    32x &+   16 \\
   (x+3)^5 = &\star x^5 &+ \star x^4 &+ \star x^3 &+ 270x^2 &+ 405x &+ 243
\end{array}
\begin{array}{c|c}
   \times &16 & 32x &24x^2 &\cdots \\
\hline
   243 & \star & \star x & 5832x^2 &\cdots \\
   405x & \star x & 12960x^2 & \star x^3 &\cdots \\
   270x^2 & 4320 x^2 & \star x^3 & \star x^4 &\cdots \\
   \vdots & \vdots & \vdots& \vdots  \\
\hline
\end{array}
A: Wouldn't the $x$ term of one and the $x$ term of the other combine in the product to produce an $x^2$ term in the result?
