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Suppose the area of a rectangle is 1023 +37x2 + 38x + 20 and the length is 2x + 5. What is the width of the rectangle?Width:

Suppose the area of a rectangle is 1023 +37x2 + 38x + 20 and the length is 2x + 5. What-example-1
User Jaime Sangcap
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1 Answer

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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}

For our rectangle, this equation looks like


10x^3+37x^2+38x+20=(2x+5)\cdot W\text{.}

Solving this equation for W, we get


W=(10x^3+37x^2+38x+20)/(2x+5)\text{.}

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.

And Thus,


W=5x^2+6x+4.Answer

The width of the given rectangle is


5x^2+6x+4.

Suppose the area of a rectangle is 1023 +37x2 + 38x + 20 and the length is 2x + 5. What-example-1
Suppose the area of a rectangle is 1023 +37x2 + 38x + 20 and the length is 2x + 5. What-example-2
User Kevin Tianyu Xu
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3.2k points