Verified Answer :

  • The term \(5mn\) must be a factor of any monomial that has a GCF of \(5mn\) with another term.
  • Check each term for divisibility by 5 and the presence of both variables \(m\) and \(n\).
  • Terms \(m^5 n^5\) and \(24m^3 n^4\) fail the coefficient divisibility check.
  • The terms \(5m^4 n^3\), \(10m^4 n\), and \(15m^2 n^2\) are divisible by \(5mn\) and qualify as having a GCF of \(5mn\) with other monomials.
  • Final answer: \(\boxed{5m^4 n^3, 10m^4 n, 15m^2 n^2}\)

Explanation

  1. Analyzing the Condition for GCF

To determine which of the given terms could have a greatest common factor (GCF) of \(5mn\) with another monomial, we must first recognize that if a monomial \(T\) and some other monomial \(S\) have a GCF of \(5mn\), then \(5mn\) must be a factor of \(T\). If \(5mn\) is not a factor of \(T\), then the GCF of \(T\) and any \(S\) cannot be \(5mn\), because the GCF must be a common divisor.

Therefore, our primary task is to identify which of the provided expressions are divisible by \(5mn\). An expression is divisible by \(5mn\) if its numerical coefficient is a multiple of 5 and it contains at least one power of \(m\) and at least one power of \(n\).

  1. Checking Each Term

Let us evaluate each term:

  1. \(m^5 n^5\): The coefficient is 1, which is not divisible by 5. Thus, \(5mn\) is not a factor.
  2. \(5m^4 n^3\): The coefficient is 5, which is divisible by 5. It contains \(m^4\) and \(n^3\). Thus, \(5mn\) is a factor: \(5m^4 n^3 = (5mn) \cdot (m^3 n^2)\). Since \(m^3 n^2\) is a monomial, we can choose \(S = 5mn\), such that \(GCF(5m^4 n^3, 5mn) = 5mn\).
  3. \(10m^4 n\): The coefficient is 10, which is divisible by 5. It contains \(m^4\) and \(n^1\). Thus, \(5mn\) is a factor: \(10m^4 n = (5mn) \cdot (2m^3)\). Since \(2m^3\) is a monomial, we can choose \(S = 5mn\), such that \(GCF(10m^4 n, 5mn) = 5mn\).
  4. \(15m^2 n^2\): The coefficient is 15, which is divisible by 5. It contains \(m^2\) and \(n^2\). Thus, \(5mn\) is a factor: \(15m^2 n^2 = (5mn) \cdot (3mn)\). Since \(3mn\) is a monomial, we can choose \(S = 5mn\), such that \(GCF(15m^2 n^2, 5mn) = 5mn\).
  5. \(24m^3 n^4\): The coefficient is 24, which is not divisible by 5. Thus, \(5mn\) is not a factor.
  1. Conclusion

Based on our analysis, the terms that have \(5mn\) as a factor—and thus can form a pair with a GCF of \(5mn\)—are \(5m^4 n^3\), \(10m^4 n\), and \(15m^2 n^2\). The terms \(m^5 n^5\) and \(24m^3 n^4\) do not contain 5 as a factor in their coefficients, so they cannot have a GCF of \(5mn\). Therefore, the correct terms are \(\boxed{5m^4 n^3, 10m^4 n, 15m^2 n^2}\).

Examples

In logistics and distribution, imagine you are organizing inventory into crates. Each crate must contain a specific number of items, represented by monomials. If you need to group different types of crates such that they share a common "baseline" unit (the GCF) of 5 metric tons of material (represented by \(5mn\)), this method helps you quickly identify which inventory bundles are compatible for this classification. By checking if each bundle is a multiple of your baseline unit, you can efficiently sort and organize your resources.