Answer
$8x^2-12x+4$.
Work Step by Step
The given area is $A=8x^3-6x^2-5x+3$
and the width is $W=x+\frac{3}{4}$.
The area of the rectangle is $A=L\cdot W$,
where the length is $L$ and the width is $W$.
Isolate $L$.
$L=\frac{A}{W}$
Substitute the expressions of $A$ and $W$.
$L=\frac{8x^3-6x^2-5x+3}{x+\frac{3}{4}}$
We can write as
$L=(8x^3-6x^2-5x+3)\div(x+\frac{3}{4})$
The value of $c$ is $-\frac{3}{4}$.
Divide the polynomial $8x^3-6x^2-5x+3$ by $x-c$, where $c=-\frac{3}{4}$, using synthetic division:
$\begin{matrix}
-\frac{3}{4}) &8&-6&-5&3 \\
& &-6&9&-3 \\
& --&--&--& --\\
& 8&-12&4&0
\end{matrix}$
The quotient is $8x^2-12x+4$ and the remainder is zero.
Hence, the length is $8x^2-12x+4$.
