Answer
$h=2x+2+\frac{\color{red}{0}}{\color{red}{}\color{blue}{x^2+x-3}}$
Work Step by Step
The formula for the area, $A$, of a trapezoid is given by
$$\begin{aligned}
A&=\frac{(b_1+b_2)h}{2}
\\
\text{where:}&
\\
b_1,b_2&=\text{ bases}
\\
h&=\text{ height}\end{aligned}
.$$
Using the formula for the area of a trapezoid, with $A=x^3+2x^2-2x-3$, $b_1=x$, and $b_2=x^2-3$, then
$$\begin{aligned}
A&=\frac{(b_1+b_2)h}{2}
\\
x^3+2x^2-2x-3&=\frac{(x+x^2-3)h}{2}
\\
\frac{2(x^3+2x^2-2x-3)}{x+x^2-3}&=h
\\
h&=\frac{2x^3+4x^2-4x-6}{x^2+x-3}
\end{aligned}
.$$
Using the long division method shown below, then
$$\begin{aligned}
&
(2x^3+4x^2-4x-6)\div(x^2+x-3)
\\&=
2x+2+\frac{\color{red}{0}}{\color{red}{}\color{blue}{x^2+x-3}}
.\end{aligned}
$$
Hence, based on the required form, an expression for the height, $h$, is
$$
h=2x+2+\frac{\color{red}{0}}{\color{red}{}\color{blue}{x^2+x-3}}
.$$
