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: 30

Answer

$500 = 2\cdot 2\cdot 5\cdot 5\cdot 5$

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 $500$ we write the number as a product of factors and continue the process until all factors are prime numbers. We start by writing $500$ as a product of two numbers: because the ones digit is even, $500$ is divisible by $2$: $500=2\cdot 250$. The number $2$ is prime, but $250$ is not. Because the ones digit is even, $250$ is divisible by $2$, so $250=2\cdot 125$: $500=2\cdot 2\cdot 125$. As $125$ is not a prime number and its ones digit is $0$ or $5$, it follows that $125$ is divisible by $5$, so $125=5\cdot 25$: $500=2\cdot 2\cdot 5\cdot 25$. As $25$ is not a prime number we write: $25=5\cdot 5$ and we have: $500=2\cdot 2\cdot 5\cdot 5\cdot 5$. Now each factor is a prime number, therefore the prime factorization of $500$ is $2\cdot 2\cdot 5\cdot 5\cdot 5$.
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