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What is the factor of 16^2-81
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What is the factor of 16^2-81

User Dvdsmpsn
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this is the diffiarence of two perfect squares. this should be memorised, but if not here it is

a^2-b^2=(a-b)(a+b)
so 16^2-81=16^2-9^2
so the answer is (16-9)(16+9)
User QuantumChris
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