The Highest Common Factor of 6b², 12b⁴ and 24abc is
AAAaannndd the bot gets it wrong yet again!
sorry, 6b^2 does not divide 24abc.
The GCF is 6b.
My apologies for the error in my previous response. You are correct that the highest common factor (HCF) of 6b², 12b⁴, and 24abc is 6b, not 6b².
To see why, we can look at the prime factorization of each term:
6b² = 2 × 3 × b × b
12b⁴ = 2² × 3 × b⁴
24abc = 2³ × 3 × a × b × c
The factors that these terms have in common are 2, 3, and b. The lowest power of 2 that appears in all three terms is 2 (from the factor of 2 in 6b² and the factor of 2² in 12b⁴). The lowest power of 3 that appears in all three terms is 1. The lowest power of b that appears in all three terms is 1 (from the factor of b in 6b² and the factor of b in 12b⁴). Therefore, the HCF is 2 × 3 × b = 6b.
Once again, I apologize for any confusion caused by my previous response.
Good oobleck got it :)
Thank you for your understanding, I strive to provide the correct answers to the best of my abilities. Let me know if you have any further math questions!
To find the Highest Common Factor (HCF) of 6b², 12b⁴, and 24abc, we need to identify the common factors of all the given terms.
Step 1: Factorize each term individually.
- For 6b², we can break it down into 2 * 3 * b * b.
- For 12b⁴, we have 2 * 2 * 3 * b * b * b * b.
- For 24abc, we can factor it as 2 * 2 * 2 * 3 * a * b * c.
Step 2: Identify the common factors.
When comparing the factorization of all the terms, we can see that the common factors are 2, 3, and b. The exponents of b are 1, 2, and 1 in each term respectively, so the HCF for b is b^1.
Step 3: Calculate the HCF.
Multiplying the common factors together, we get HCF = 2 * 3 * b^1 = 6b.
Therefore, the HCF of 6b², 12b⁴, and 24abc is 6b.
6b².
To find the highest common factor (HCF) of these three terms, we need to find the highest power of each factor that appears in all three terms.
For 6b², the prime factors are 2 and 3, and the highest power of 2 that appears is 1 (since 6 = 2 x 3 and there is no other factor of 2), while the highest power of 3 that appears is also 1. There is only one factor of b², so the highest power of b² that appears is 2. Therefore, the prime factorization of 6b² is 2 x 3 x b².
For 12b⁴, the prime factors are 2 and b, and the highest power of 2 that appears is 2 (since 12 = 2² x 3), while the highest power of b that appears is 4. Therefore, the prime factorization of 12b⁴ is 2² x b⁴.
For 24abc, the prime factors are 2, 3, a, and b, and the highest power of 2 that appears is 3 (since 24 = 2³ x 3), while the highest power of 3 that appears is 1. There is one factor of a and one factor of b, so the highest powers of a and b that appear are both 1. Therefore, the prime factorization of 24abc is 2³ x 3 x a x b.
To find the HCF, we need to take the product of the lowest power of each common factor. In this case, the only common factor is 2 x b². The lowest power of 2 x b² that appears in all three terms is 6b², since that is the product of the lowest powers of 2 and b² that appears in each term. Therefore, the HCF is 6b².