Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1 - Introduction to Algebraic Expressions - 1.3 Fraction Notation - 1.3 Exercise Set: 30

Answer

$2^{3}\times7$

Work Step by Step

A number is prime if it has only two factors: 1 and itself. We know that the factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56. 2 is a prime number, so we can divide 56 by 2 $56\div2=28$ The factors of 28 are 1, 2, 4, 7, 14, and 28. Since 2 is a prime number, divide by 2. $28\div2=14$ The factors of 14 are 1, 2, 7, and 14. Since 2 is a prime number, divide by 2. $14\div2=7$ 7 is also a prime number Therefore, the prime factorization of 56 is $2\times2\times2\times7=2^{3}\times7$.
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