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: 59

Answer

a) $40 = 2\cdot 2 \cdot 2 \cdot 5$ b) $40 = 2\cdot 2 \cdot 2 \cdot 5$ c) Similarities: same prime factorization. Differences: the order in which the prime factors are determined.

Work Step by Step

a) To write the prime factorization of $40$ using $2$ and $20$ as the first pair of factors, we can start by expressing $20$ as a product of its prime factors: $20 = 2 \cdot 2 \cdot 5$ Then, we can write $40$ as the product of $2$ (from the first pair of factors) and the prime factorization of $20$: $40 = 2\cdot (2\cdot 2 \cdot 5)$ Simplifying further, we get: $40 = 2\cdot 2\cdot 2\cdot 5$ b) To write the prime factorization of $40$ using $4$ and $10$ as the first pair of factors, we need to express $4$ and $10$ as products of their prime factors: $4 = 2 \cdot 2$ $10 = 2 \cdot 5$ Then, we can write $40$ as the product of $4$ (from the first pair of factors) and the prime factorization of $10$: $40 = (2 \cdot 2) \cdot (2 \cdot 5)$ Simplifying further, we get: $40 = 2\cdot 2\cdot 2 \cdot 5$ c) In both parts a) and b), the prime factorization of $40$ is expressed as $2\cdot 2\cdot 2\cdot 5$. The difference lies in the arrangement of the factors. In part a), the first pair of factors used is $2$ and $20$. The factorization starts with $20$, which is then broken down into its prime factors ($2 \cdot 2 \cdot 5$). This is followed by multiplying $2$ (from the first pair) with the prime factorization of $20$. In part b), the first pair of factors used is $4$ and $10$. The factorization starts with $4$ (already expressed as a product of its prime factors, $2 \cdot 2$), and then the prime factorization of $10$ ($2\cdot 5$) is multiplied with it. The differences lie in the arrangement and the way the factorization is presented, depending on the initial pair of factors used.
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