1.2 Prime factorization

Find the prime factorization of 10.

$10=2·5$

Both 2 and 5 are prime numbers. Thus, $2·5$ is the prime factorization of 10.

Find the prime factorization of 60.

$\begin{array}{lllll}60\hfill & =\hfill & 2·30\hfill & \hfill & \text{30\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.30=2·15\hfill \\ \hfill & =\hfill & 2·2·15\hfill & \hfill & 15\text{\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.15=3·5\hfill \\ \hfill & =\hfill & 2·2·3·5\hfill & \hfill & \text{We'll\hspace{0.17em}use\hspace{0.17em}exponents}\text{.\hspace{0.17em}}2·2=2^2\hfill \\ \hfill & =\hfill & 2^2·3·5\hfill & \hfill & \hfill \end{array}$

The numbers 2, 3, and 5 are all primes. Thus, $2^2·3·5$ is the prime factorization of 60.

Find the prime factorization of 11.

11 is a prime number. Prime factorization applies only to composite numbers.

Find the prime factorization of 60.

Since 60 is an even number, it is divisible by 2. We will repeatedly divide by 2 until we no longer can (when we start getting a remainder). We shall divide in the following way.

   $\begin{array}{l}\text{30\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2\hspace{0.17em}again}\text{.}\hfill \\ \text{15\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{5\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}5,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}\text{.}\hfill \end{array}$

The prime factorization of 60 is the product of all these divisors.

$\begin{array}{lllll}60\hfill & =\hfill & 2·2·3·5\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 60\hfill & =\hfill & 2^2·3·5\hfill & \hfill & \hfill \end{array}$

Find the prime factorization of 441.

Since 441 is an odd number, it is not divisible by 2. We’ll try 3, the next largest prime.

   $\begin{array}{l}\text{147\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3}.\hfill \\ \text{49\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3\hspace{0.17em}nor\hspace{0.17em}by\hspace{0.17em}5},\text{\hspace{0.17em}but\hspace{0.17em}by\hspace{0.17em}7}.\hfill \\ \text{7\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}7}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}.\hfill \end{array}$

The prime factorization of 441 is the product of all the divisors.

$\begin{array}{lllll}441\hfill & =\hfill & 3·3·7·7\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 441\hfill & =\hfill & 3^2·7^2\hfill & \hfill & \hfill \end{array}$