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

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Answer 1 · 2016-12-09T02:57:49

#LCM=2xx2xx2xx2xx5xx7xxaxxbxxbxxbxxb=560ab^4#

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#