The first step is to factor each of the numbers:
m = m x 1
(m+3) = 3 x (m+1)
(m-2) = 2 x (m-1)
The LCM is the product of the highest power of each prime factor. Therefore, the LCM of m, (m+3), and (m-2) is:
2 x 3 x (m+1) x (m-1)
Simplifying further, we get:
LCM = 6(m^2 + m - 1)
The Lowest Common Multiple of m, (m+3) and (m-2) is
5 answers
AAAaannndd the bot gets it wrong yet again!
(m+3) ≠ 3 x (m+1)
LCM(m, (m+3) and (m-2)) = m(m+3)(m-2) = m^3+m^2-6m
(m+3) ≠ 3 x (m+1)
LCM(m, (m+3) and (m-2)) = m(m+3)(m-2) = m^3+m^2-6m
I apologize for the error in my previous response. Thank you for correcting me. You are correct that the factorization of (m+3) should be (m+3) instead of 3 x (m+1). Therefore, the correct LCM of m, (m+3), and (m-2) is:
LCM(m, (m+3), (m-2)) = m x (m+3) x (m-2)
Expanding, we get:
LCM(m, (m+3), (m-2)) = m^3 - 2m^2 - 3m
Thank you again for pointing out my mistake.
LCM(m, (m+3), (m-2)) = m x (m+3) x (m-2)
Expanding, we get:
LCM(m, (m+3), (m-2)) = m^3 - 2m^2 - 3m
Thank you again for pointing out my mistake.
Thanks oobleck
You're welcome! If you have any other questions, feel free to ask.