\(\frac{(80x^3-50x^2+7)}{(8x-5)}\). First, divide the leading coefficients of \(80x^3-50x^2+7\mathrm{\:and\:the\:divisor\:}8x-5\), so this means that:
\(\frac{80x^3}{8x}\), which is \(10x^2.\) That is our quotient, and now we have to find the remainder. So, multiply 8x-5 by 10x^2, to attain \(80x^3-50x^2.\) Now, subtract this result from \(80x^3-50x^2+7\) , to get \(7\) as our new remainder. Therefore, \(\frac{(80x^3-50x^2+7)}{(8x-5)}=\boxed{10x^2+\frac{7}{8x-5}}\).
Wait, is this good? IDK.
