Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter R - Section R.1 - Factors and the Least Common Multiple - Exercise Set - Page R-7: 29

Answer

$300 = 2\cdot 2\cdot 3\cdot 5\cdot 5$

Work Step by Step

The $\textit{prime factorization}$ of a number is obtained by writing the number as a product of primes. To determine the prime factorization of $300$ we write the number as a product of factors and continue the process until all factors are prime numbers. We start by writing $300$ as a product of two numbers: because the ones digit is even, $300$ is divisible by $2$: $300=2\cdot 150$. The number $2$ is prime, but $150$ is not. Because the ones digit is even, $150$ is divisible by $2$, so $150=2\cdot 75$: $300=2\cdot 2\cdot 75$. As $75$ is not a prime number and the sum of its digits is $7+5=12$ divisible by $3$, it follows that $75$ is divisible by $3$, so $75=3\cdot 25$: $300=2\cdot 2\cdot 3\cdot 25$. As $25$ is not a prime number we write: $25=5\cdot 5$ and we have: $300=2\cdot 2\cdot 3\cdot 5\cdot 5$. Now each factor is a prime number, therefore the prime factorization of $300$ is $2\cdot 2\cdot 3\cdot 5\cdot 5$.
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