The Lowest Common Multiple of m, (m+3) and (m-2) is
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
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.
Thanks oobleck
You're welcome! If you have any other questions, feel free to ask.
To find the Lowest Common Multiple (LCM) of three numbers m, (m+3), and (m-2), we need to find the smallest multiple that is divisible by all three numbers.
One way to find the LCM is by factoring the numbers and then multiplying the highest power of each factor. Let's go through the steps:
Step 1: Factor the three numbers:
m has no factors other than itself.
(m+3) can be expressed as (m+3).
(m-2) can be expressed as (m-2).
Step 2: Identify the highest power of each factor:
Since each number has only one factor, there are no highest powers to consider.
Step 3: Multiply the factors:
(m)(m+3)(m-2)
Therefore, the LCM of m, (m+3), and (m-2) is simply (m)(m+3)(m-2).
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)