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Find the GCF of the group of monomials

9a^2b^3 and 15ab^2 and 21a^4b^3

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Answer:

3ab^2

Explanation:

The GCF of 9, 15, and 21 is 3.

The GCF of a^2b^3, ab^2, and a^4b^3 is ab^2.

The answer is the product of the two terms which is 3ab^2.

User PintoDoido
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