​ Synthetic division

$\frac{(80x^3-50x^2+7)}{(8x-5)}$. 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.

Set 8x  - 5  = 0  ⇒    x =  5/8....this is what we need to divide by

Note that the polynomial is really   80x^3  - 50x^2  + 0x   + 7

5/8  [  80      -50       0        7  ]

                     50       0        0

         _____________________

         80         0        0         7

The remainder is correct

The apparent remaining polynomial is  80x^2

Note....RP...that all we really need to do to find the correct remaining polynomial is just to divide  the apparent remaining polynomial  by   the "a"  coefiicient  of the  divisor, ax - b.....in this case.....ax - b  = 8x  - 5...so we can divide

80x^2  by   8  = 10x^2  and this is the correct residual polynomial

So...the answer   10x^2 R [ 7 /(8x - 5) ] 

Does that help????