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when 8x^-3-64x is factored completely, one of the factors is?
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=8x(x^2-8) so either 8x or x^2 -8
thanks , man you're good at this lol good job(: i wish i had that capability
just think of what they have in common both 8 and 64 are divisible by 8 and they both have an x anyway thanks
so how you get x^2-8?
by dividing each term by 8x \[\frac{8x^{3}}{8x} = x^{2}\] \[\frac{64x}{8x} = 8\]
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