Answer
$(2b^2+b+3)\text{ in.}$
Work Step by Step
Using $A=lw$ or the formula for the area, $A$, of a rectangle with length $(l)$ and width $(w)$, then
$$\begin{aligned}
A&=lw
\\
4b^3+5b-3&=l(2b-1)
\\
l&=\frac{4b^3+5b-3}{2b-1}
.\end{aligned}$$
Using long division, then
$$\begin{array}{l}
\phantom{d+3)^2}2b^2+\phantom{2}b+\phantom{5}3
\\
\color{blue}{2b-1}\color{black}{\overline{\big)4b^3\phantom{-2b^2}+5b-3}}
\\
\phantom{2b-1)}\underline{4b^3-2b^2}
\\
\phantom{2b-1)4b^3-\,}2b^2+5b
\\
\phantom{2b-1)4b^3-}\underline{\,2b^2-\phantom{5}b}
\\
\phantom{2b-1)4b^3-2b^2+}6b-3
\\
\phantom{2b-1)4b^3-2b^2+}\underline{6b-3}
\\
\color{red}
{\phantom{2b-1)4b^3-2b^2+4b-}0}
\end{array}$$Hence, the length of the rectangle is
$$
(2b^2+b+3)\text{ in.}
$$
