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how do I factor 81c^2-16
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what number squared is 16?
4 right
right and what would you square to get \(81c^2\) ?
umm 9c^2
well actually \(9c\)
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since \((9c)^2=81c^2\)
now factor the difference of two squares as \[a^2-b^2=(a+b)(a-b)\] with \(a=9c\) and \(b=4\)
is it 65 = (9+4) (9-4) right
there is no 65 in the question, and you are missing a c
\[81c^2-16=(9c+4)(9c-4)\]
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what do I do now I am lost
hello there
help me please
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