Answer
$0.5x^2-0.4x+0.3$.
Work Step by Step
The given area is $A=0.5x^3-0.3x^2+0.22x+0.06$.
and the width is $W=x+0.2$.
The area of the rectangle $A=L\cdot W$,
where the length is $L$ and the width is $W$.
Isolate $L$.
$L=\frac{A}{W}$
Substitute the expressions for $A$ and $W$:
$L=\frac{0.5x^3-0.3x^2+0.22x+0.06}{x+0.2}$
We can write as
$L=(0.5x^3-0.3x^2+0.22x+0.06)\div(x+0.2)$
We divide the polynomial $0.5x^3-0.3x^2+0.22x+0.06$ by $x-c$, where $c=-0.2$, using synthetic division:
$\begin{matrix}
-0.2) &0.5&-0.3&0.22&0.06 \\
& &-0.10&0.08&-0.06 \\
& --&--&--& --\\
& 0.5&-0.40&0.30&0
\end{matrix}$
The quotient is $0.5x^2-0.4x+0.3$ and the remainder is zero.
Hence, the length is $0.5x^2-0.4x+0.3$.
