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Divide. x^2-4/x-8/x-2
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\[\frac{\frac{x^2-4}{x-8}}{x-2}\]?
yes.
same as \[\frac{x^2-4}{(x-8)(x-2)}\] factor the numerator, and cancel the common factor of \(x-2\) top and bottom
as in \[\frac{(x+2)(x-2)}{(x-8)(x-2)}\] then cancel
so x+2/x-8?
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yes
ok ty satellite73
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