Q1 A of 42 Page 136

Let’s find G.C.D of the following algebraic expressions:

4a2b2, 20ab2


4a2b2, 20ab2


Factors of 4a2b2=4×a×a×b×b


Factors of 20ab2=4×5×a×b×b


GCD is the greatest common divisor, which is equal to the product of all the common divisors.


The GCD of 4a2b2, 20ab2 is 4ab2


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