How do you find the Least common multiple of 28b^2, 20ab^3, 16b^4?

1 Answer

LCM=2xx2xx2xx2xx5xx7xxaxxbxxbxxbxxb=560ab^4

Explanation:

To find LCM, I like first to do a prime factorization of the numbers in question:

28b^2=2xx14xxbxxb=2xx2xx7xxbxxb

20ab^3=2xx10xxaxxbxxbxxb=2xx2xx5xxaxxbxxbxxb

16b^4=2xx8xxbxxbxxbxxb=2xx2xx4xxbxxbxxbxxb=2xx2xx2xx2xxbxxbxxbxxb

The LCM will include all the primes listed (and all the variables too) in the biggest quantity from any given number we're working with.

For instance, we have 2s. There are 4 of them in the 16b^4, so our LCM will have 4 as well:

LCM=2xx2xx2xx2xx?

We also have a 5 in the 20ab^3:

LCM=2xx2xx2xx2xx5xx?

There's a 7 in the 28b^2

LCM=2xx2xx2xx2xx5xx7xx?

That's it for numbers. We have an a in 20ab^3 and we have 4 b in 16b^4

LCM=2xx2xx2xx2xx5xx7xxaxxbxxbxxbxxb

All of this gives us:

LCM=2xx2xx2xx2xx5xx7xxaxxbxxbxxbxxb=560ab^4

~~~

28b^2xx2xx2xx5xxaxxbxxb=560ab^4
20ab^3xx2xx2xx7xxb=560ab^4
16b^4xx5xx7xxa=560ab^4