Answer
$ 2b^2 +b+3$.
Work Step by Step
The given values are
Width $w=(2b-1)$.
Area $A=(4b^3+5b-3) \; in.^2$
Formula for area is $l\times w=A$.
Therefore, $(4b^3+5b-3)=l(2b-1)$.
Divide both sides by $(2b-1)$.
Therefore, $\frac{(4b^3+5b-3)}{(2b-1)}=l$.
Rewrite the expression in standard form.
$(4b^3+0b^2+5b-3)\div(2b-1)$
$\begin{matrix}
& 2b^2 & +b &+3 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
2b-1) &4b^3&+0b^2&+5b&-3 & \\
& 4b^3 & -2b^2 & & & \leftarrow &2b^2(2b-1) \\
& -- & -- & & & \leftarrow &subtract \\
& 0 & 2b^2 & +5b & & \\
& & 2b^2 & -b & & \leftarrow & b(2b-1) \\
& & -- & -- & & \leftarrow & subtract \\
& & 0&6b &-3 & \\
& & & 6b& -3 & \leftarrow & 3(2b-1) \\
& & & -- & -- & \leftarrow & subtract \\
& & & 0 & 0 & \leftarrow & Remainder
\end{matrix}$
The answer is
$\Rightarrow Quotient + \frac{Remainder}{Divisor}$
$\Rightarrow 2b^2 +b+3+\frac{0}{2b-1}$
Simplify.
$\Rightarrow 2b^2 +b+3$.
