Question:
Find the greatest common factor of the terms in each of the following expression:
3a2b2 + 4b2c2 + 12a2b2c2
Solution:
The expression has three monomials: 3a2b2, 4b2c2 and 12a2b2c2.
The numerical coefficients of the given monomials are 3, 4 and 12. The greatest common factor of 3, 4 and 12 is 1.
The common literal appearing in the three monomials is b.
The smallest power of b in the three monomials is 2.
The monomial of common literals with the smallest powers is b2.
Hence, the greatest common factor is b2.