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completely simplify (x-2)/(8x^2- 32)
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\[\frac{x-2}{8x^2-32}\]
(x-2)/(8(x^2-4))
(x-2)/(8(x+2)(x-2)) \[\frac{ (x-2) }{ (8(x+2)(x-2)) }\]
first factor out an 8 out of the bottom \[\frac{x-2}{8(x^2-4)}\] next use \[(a-b)^2 =(a-b)(a+b)\] when a and b are perfect squares
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\[a^2-b^2=(a+b)(a-b)\]
ok
1/8(x+2) is this right?
= (x-2)/8(x^2-4) = (x-2)/8(x+2)(x-2) = 1/8(x+2)
yes thats absolutely corret:)
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