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: 32

Answer

$2^{2}\times3^{2}\times5$

Work Step by Step

A number is prime if it has only two factors: 1 and itself. We know that the factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, and 180. 2 is a prime number, so we can divide 180 by 2 $180\div2=90$ The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90. Since 2 is a prime number, divide by 2. $90\div2=45$ The factors of 45 are 1, 3, 9, 15 and 45. Since 3 is a prime number, divide by 3. $45\div3=15$ The factors of 15 are 1, 3, 5, and 15. Since 3 is a prime number, divide by 3. $15\div3=5$ 5 is also a prime number Therefore, the prime factorization of 180 is $2\times2\times3\times3\times5=2^{2}\times3^{2}\times5$.
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