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.
The Highest Common Factor of 6b², 12b⁴ and 24abc is
5 answers
AAAaannndd the bot gets it wrong yet again!
sorry, 6b^2 does not divide 24abc.
The GCF is 6b.
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.
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!