Answer
$2b^2+\frac{1}{4}b+\frac{29}{32}+\frac{{29}}{{32}{(8b-1)}}
$
Work Step by Step
Re-arranging the order of the terms and leaving space for missing terms, the given is equivalent to
$$
(16b^3+0b^2+7b+0)÷(8b-1)
.$$
Using the long division method shown below, then
$$\begin{aligned}
&
(16b^3+7b)÷(8b-1)
\\&=
2b^2+\frac{1}{4}b+\frac{29}{32}+\frac{\color{red}{29}}{\color{red}{32}\color{blue}{(8b-1)}}
.\end{aligned}
$$
