What is the GCF and LCM for #22xy^2z^2#, #33x^2yz^2#, #44x^2yz#?

Question

What is the GCF and LCM for  #22xy^2z^2#, #33x^2yz^2#, #44x^2yz#?

Eva Adamson · Accepted Answer

GCF: #11xyz#
LCM: #132x^2y^2z^2# GCF: Basically we find the things that all things has in common. For this one, we can see that all of them have at least one #x#, one #y# and one #z#, so we can say that #xyz# is a factor, dividing them all by it, we get #22yz#, #33xz# and #44x# Now, remember that #22 = 11*2#, #33 = 11*3# and #44 = 11*4#, so we can say that 11 is also a common factor Dividing them all by #11xyz# we get #2yz#, #3xz# and #4x# There is no more we can factor out, the GCF is #11xyz# LCM: Basically we want the smallest term we can get that is a multiple of all three of these terms, i.e.: the smallest non-zero number (or monomial) that is perfectly divisible by all three terms. We separate the variables and constants to make our life easier, so we need to find the LCM of 22, 33 and 44, so by the rules of that (divide by the smallest prime and work up) #22, 33, 44 | 2# #11, 33, 22 | 2# #11, 33, 11| 3# #11, 11, 11| 11# #color(white)(0)1,color(white)(0)1,color(white)(0)1|2^2*3*11=12*11 = 132# And the LCM of #xy^2z^2#, #x^2yz^2# and #x^2yz#, using the same rules, but now we assume that each variable is a prime number. #xy^2z^2, x^2yz^2, x^2yz | x# #color(white)(x)y^2z^2, x^color(white)(2)yz^2, x^color(white)(2)yz | x# #color(white)(x)y^2z^2, color(white)(x^2)yz^2, color(white)(x^2)yz|y# #color(white)(x)y^color(white)(2)z^2, color(white)(x^2y)z^2, color(white)(x^2y)z|y# #color(white)(xy^2)z^2, color(white)(x^2y)z^2, color(white)(x^2y)z|z# #color(white)(xy^2)z^color(white)(2), color(white)(x^2y)z^color(white)(2), color(white)(x^2y)1|z# #color(white)(xy^2)1^color(white)(2), color(white)(x^2y)1^color(white)(2), color(white)(x^2y)1|x^2*y^2*z^2# Multiply the two together to find the LCM, which is #132x^2y^2z^2#

Hazel Anglin · Answer

undefined The GCF is 11xyz and the LCM is 132x^2y^2z^2.