$(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$ and $\frac{A_{19}}{A_5}=\frac{1}{2^7},$ then $c$ is? $(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$. The problem is that we can't express $(3x^2+2x+c)^{12}$ as a perfect square
There was a hint in my book, put $x=\frac{c}{3x}$ but that isn't leading anywhere either
Let $\displaystyle x=\frac{c}{3x}$ Then $\displaystyle\left(\frac{c^2}{3x^2}+\frac{2c}{3x}+c\right)^{12}*\frac{x^{24}}{x^{24}}=\left(\frac{x^2+\frac{2cx}{3}+\frac{c^2}{3}}{x^2}\right)^{12}$
We can also do this through the general term of multinomial coefficient, but that would be a very long method and I only have $5$ minutes in the exam. Can someone help, thanks $:)$

Let $x\rightarrow\frac{c}{3x}$
$$(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$$
$$\left(\frac{c^2}{3x^2}+\frac{2c}{3x}+c\right)^{12}=\sum_{r=0}^{24}A_rc^r3^{-r}x^{-r}$$
$$\frac{c^{12}}{3^{12}x^{24}}\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r}3^{-r}x^{-r}$$
$$\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r-12}3^{12-r}x^{24-r}$$
Then the coefficient of $x^5$ on both sides should be the same which gives
$$A_5=A_{19}c^{7}3^{-7}\implies c^{7}=3^{7}\left(\frac{A_{19}}{A_5}\right)^{-1}\implies c=6$$
