Step-by-step explanation
Let's picture the situation of the exercise:
We're looking for the width of the above rectangle. Recall that the area of a rectangle is given by the following equation:
![A=L\cdot W\leftarrow\begin{cases}A=area, \\ L=\text{length,} \\ W=\text{width.}\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/qrx59x6qegwpybi5iqws0ygttyfs822ic5.png)
For our rectangle, this equation looks like
![10x^3+37x^2+38x+20=(2x+5)\cdot W\text{.}](https://img.qammunity.org/2023/formulas/mathematics/college/odm72k9uozkvsipy21jukumorfmi3n08m9.png)
Solving this equation for W, we get
![W=(10x^3+37x^2+38x+20)/(2x+5)\text{.}](https://img.qammunity.org/2023/formulas/mathematics/college/baij96n6felgx86307bk7nmzc2md08mp7z.png)
Then, the exercise turns out to be a polynomial division. let's do it:
Then,
![(10x^3+37x^2+38x+20)/(2x+5)=\text{ Quotient}+\text{ Residue}=(5x^2+6x+4)+0=5x^2+6x+4.](https://img.qammunity.org/2023/formulas/mathematics/college/no8m9mlj8rdphp5ikke1kfzzfvgyts50lp.png)
And Thus,
![W=5x^2+6x+4.](https://img.qammunity.org/2023/formulas/mathematics/college/rqifowfmkejrtu4032t8qnro01ji6jc5d3.png)
Answer
The width of the given rectangle is
![5x^2+6x+4.](https://img.qammunity.org/2023/formulas/mathematics/college/byxyfdbuzx5lort60ac89legcwepu80c89.png)