Expand $$$\left(2 x + 3\right)^{4}$$$

Expand $$$\left(2 x + 3\right)^{4}$$$.

The expansion is given by the following formula: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, where $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ and $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

We have that $$$a = 2 x$$$, $$$b = 3$$$, and $$$n = 4$$$.

Therefore, $$$\left(2 x + 3\right)^{4} = \sum_{k=0}^{4} {\binom{4}{k}} \left(2 x\right)^{4 - k} 3^{k}$$$.

Now, calculate the product for every value of $$$k$$$ from $$$0$$$ to $$$4$$$.

$$$k = 0$$$: $$${\binom{4}{0}} \left(2 x\right)^{4 - 0} \cdot 3^{0} = \frac{4!}{\left(4 - 0\right)! 0!} \left(2 x\right)^{4 - 0} \cdot 3^{0} = 16 x^{4}$$$

$$$k = 1$$$: $$${\binom{4}{1}} \left(2 x\right)^{4 - 1} \cdot 3^{1} = \frac{4!}{\left(4 - 1\right)! 1!} \left(2 x\right)^{4 - 1} \cdot 3^{1} = 96 x^{3}$$$

$$$k = 2$$$: $$${\binom{4}{2}} \left(2 x\right)^{4 - 2} \cdot 3^{2} = \frac{4!}{\left(4 - 2\right)! 2!} \left(2 x\right)^{4 - 2} \cdot 3^{2} = 216 x^{2}$$$

$$$k = 3$$$: $$${\binom{4}{3}} \left(2 x\right)^{4 - 3} \cdot 3^{3} = \frac{4!}{\left(4 - 3\right)! 3!} \left(2 x\right)^{4 - 3} \cdot 3^{3} = 216 x$$$

$$$k = 4$$$: $$${\binom{4}{4}} \left(2 x\right)^{4 - 4} \cdot 3^{4} = \frac{4!}{\left(4 - 4\right)! 4!} \left(2 x\right)^{4 - 4} \cdot 3^{4} = 81$$$

Thus, $$$\left(2 x + 3\right)^{4} = 16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81$$$.

$$$\left(2 x + 3\right)^{4} = 16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81$$$A