Question

find the greatest common factor.
4v⁴, v³, 12v⁴
write your answer as a constant times a product of single variables raised to exponents.

Explanation:

Step1: Find GCF of coefficients

Coefficients are 4, 1, 12. GCF of 4,1,12 is 1.

Step2: Find GCF of variables

Variables: \(v^4\), \(v^3\), \(v^4\) (wait, original terms: \(4v^4\), \(v^3\), \(12v^4\)). For \(v\) terms: lowest exponent is 3 (from \(v^3\)). So variable part is \(v^3\).

Step3: Combine

Multiply GCF of coefficients (1) and GCF of variables (\(v^3\)): \(1\times v^3 = v^3\). Wait, wait, let's recheck coefficients: 4,1,12. Wait, 1 is also a coefficient? Wait, the terms are \(4v^4\), \(v^3\) (which is \(1v^3\)), \(12v^4\). So coefficients: 4,1,12. GCF of 4,1,12: factors of 4: 1,2,4; factors of 1:1; factors of 12:1,2,3,4,6,12. Common factor is 1. Variables: \(v^4\), \(v^3\), \(v^4\). The lowest power of v is 3 (from \(v^3\)). So GCF is \(1\times v^3 = v^3\)? Wait, no, wait: wait, \(4v^4 = 4\times v\times v\times v\times v\), \(v^3 = 1\times v\times v\times v\), \(12v^4 = 12\times v\times v\times v\times v\). So the common factors: for coefficients, 4,1,12: GCF is 1. For variables, the common part is \(v^3\) (since each has at least \(v^3\)). Wait, but wait, is there a mistake? Wait, 4,1,12: GCF is 1? Wait, 1 is a factor of all, but maybe I missed. Wait, 4 and 12 have GCF 4, but 1 is also there. So when there's a 1, the GCF of coefficients is 1. So the GCF of the three terms is \(1\times v^3 = v^3\)? Wait, but let's check: \(4v^4 = v^3\times 4v\), \(v^3 = v^3\times 1\), \(12v^4 = v^3\times 12v\). Yes, that works. So the GCF is \(v^3\). Wait, but wait, maybe I made a mistake with coefficients. Wait, 4,1,12: GCF is 1? Because 1 is a factor of all, but 4 and 12 have GCF 4, but 1 is in the mix. So when one of the coefficients is 1, the GCF of coefficients is 1. So yes, the GCF is \(v^3\).

Wait, no, wait: let's list the terms:

Term 1: \(4v^4 = 4 \times v^4\)

Term 2: \(v^3 = 1 \times v^3\)

Term 3: \(12v^4 = 12 \times v^4\)

So to find GCF, we take the minimum coefficient GCF and minimum variable exponent.

Coefficients: 4, 1, 12. GCF of 4,1,12: as 1 is a coefficient, the GCF is 1 (since 1 divides 4 and 12).

Variables: \(v^4\), \(v^3\), \(v^4\). The minimum exponent of v is 3 (from \(v^3\)).

So GCF is \(1 \times v^3 = v^3\).

Answer:

\(v^3\)