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: 31

Answer

$588 = 2\cdot 2\cdot 3\cdot 7\cdot 7$

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 $588$ we write the number as a product of factors and continue the process until all factors are prime numbers. We start by writing $588$ as a product of two numbers: because the ones digit is even, $588$ is divisible by $2$: $588=2\cdot 294$. The number $2$ is prime, but $294$ is not. Because the ones digit is even, $294$ is divisible by $2$, so $294=2\cdot 147$: $588=2\cdot 2\cdot 147$. As $147$ is not a prime number and the sum of its digits is $1+4+7=12$ divisible by $3$, it follows that $147$ is divisible by $3$, so $147=3\cdot 49$: $588=2\cdot 2\cdot 3\cdot 49$. As $49$ is not a prime number we write: $49=7\cdot 7$ and we have: $588=2\cdot 2\cdot 3\cdot 7\cdot 7$. Now each factor is a prime number, therefore the prime factorization of $588$ is $2\cdot 2\cdot 3\cdot 7\cdot 7$.
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