Expand $$$\left(x - 2\right)^{5}$$$

Expand $$$\left(x - 2\right)^{5}$$$.

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 $$$a = x$$$, $$$b = -2$$$, and $$$n = 5$$$.

Therefore, $$$\left(x - 2\right)^{5} = \sum_{k=0}^{5} {\binom{5}{k}} x^{5 - k} \left(-2\right)^{k}$$$.

Now, calculate the product for every value of $$$k$$$ from $$$0$$$ to $$$5$$$.

$$$k = 0$$$: $$${\binom{5}{0}} x^{5 - 0} \left(-2\right)^{0} = \frac{5!}{\left(5 - 0\right)! 0!} x^{5 - 0} \left(-2\right)^{0} = x^{5}$$$

$$$k = 1$$$: $$${\binom{5}{1}} x^{5 - 1} \left(-2\right)^{1} = \frac{5!}{\left(5 - 1\right)! 1!} x^{5 - 1} \left(-2\right)^{1} = - 10 x^{4}$$$

$$$k = 2$$$: $$${\binom{5}{2}} x^{5 - 2} \left(-2\right)^{2} = \frac{5!}{\left(5 - 2\right)! 2!} x^{5 - 2} \left(-2\right)^{2} = 40 x^{3}$$$

$$$k = 3$$$: $$${\binom{5}{3}} x^{5 - 3} \left(-2\right)^{3} = \frac{5!}{\left(5 - 3\right)! 3!} x^{5 - 3} \left(-2\right)^{3} = - 80 x^{2}$$$

$$$k = 4$$$: $$${\binom{5}{4}} x^{5 - 4} \left(-2\right)^{4} = \frac{5!}{\left(5 - 4\right)! 4!} x^{5 - 4} \left(-2\right)^{4} = 80 x$$$

$$$k = 5$$$: $$${\binom{5}{5}} x^{5 - 5} \left(-2\right)^{5} = \frac{5!}{\left(5 - 5\right)! 5!} x^{5 - 5} \left(-2\right)^{5} = -32$$$

Thus, $$$\left(x - 2\right)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32$$$.

$$$\left(x - 2\right)^{5} = x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32$$$A