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12u^6-16u^5/2u^2 How do I do that
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\[\large \frac{u^{m}}{u^{n}} = u^{m-n}\]
But there's two numbers on top
And if you subtract them it's 4u/2u^2 and if you do that it can't divide can it?
??? what how do you get 4u ? you cant subtract they are not like terms
Oh. Well, how do you subtract the two powers if there's 3 powers in the question o:
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basic understanding of fractions it would seem... \[\frac{x+y}{2} = \frac{x}{2} + \frac{y}{2}\]
But there's still 3 powers. If there's 3 powers, how do I subtract them?
\[\frac{12u^6-16u^5}{2u^2}=\frac{4u^5(3u-4)}{2u^2} \] \[= 2u^{5-2}(3u-4) =2u^{3}(3u-4)\]
Ohhhh. So then do I just distribute?
Or is it done?
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