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divide x^8+x^4/x
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we can do x^8 + x^3. then factor that to x^2(x^4) + x
I guess that's right...
\[\frac{ x ^{a} }{ x ^{b} } =x ^{a}*x ^{-b}=x ^{(a-b)}\]
divide x^8+x^4/x = x^8 *x^-1 + x^4 *x^-1 =?
thats from the assumption divide (x^8+x^4)/x
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divide (x^8+x^4)/x = x^8 *x^-1 + x^4 *x^-1 = ? is it = x^7 + x^3 ? yes or no?
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