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Mathematics
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Factoring
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\[16^4-81\]
I know its a factor of squares but I am not sure the entire process to get it correct
\[16^4- 81= 16^4-3^4=(16^2-3^2)(16^2+3^2)\]
But notice that 16^2-3^2 is still the difference of two squares and so we can factor further:
\[(16^2-3^2)(16^2+3^2)=(16-3)(16+3)(16^2+3^2)\]
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ok and that is what I got at first but the answer showing on the exam is 2y+3
Well then I guess you have stated the problem incorrectly because it does not have a y in it.
ah yeas its 16y^4
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