Prime factorization of $$$1128$$$

Find the prime factorization of $$$1128$$$.

Start with the number $$$2$$$.

Determine whether $$$1128$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1128$$$ by $$${\color{green}2}$$$: $$$\frac{1128}{2} = {\color{red}564}$$$.

Determine whether $$$564$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$564$$$ by $$${\color{green}2}$$$: $$$\frac{564}{2} = {\color{red}282}$$$.

Determine whether $$$282$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$282$$$ by $$${\color{green}2}$$$: $$$\frac{282}{2} = {\color{red}141}$$$.

Determine whether $$$141$$$ is divisible by $$$2$$$.

Since it is not divisible, move to the next prime number.

The next prime number is $$$3$$$.

Determine whether $$$141$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$141$$$ by $$${\color{green}3}$$$: $$$\frac{141}{3} = {\color{red}47}$$$.

The prime number $$${\color{green}47}$$$ has no other factors then $$$1$$$ and $$${\color{green}47}$$$: $$$\frac{47}{47} = {\color{red}1}$$$.

Since we have obtained $$$1$$$, we are done.

Now, just count the number of occurences of the divisors (green numbers), and write down the prime factorization: $$$1128 = 2^{3} \cdot 3 \cdot 47$$$.

The prime factorization is $$$1128 = 2^{3} \cdot 3 \cdot 47$$$A.