Chapter 8.5, Problem 64PFA

To determine

To find: The width of the rectangles.

Answer to Problem 64PFA

  $\left(x^2+2\right)$

C is the correct option.

Explanation of Solution

Given:

The areas of two rectangles whose width are equal are $x^3+4x^2+2x+8\text{ and }{x}^3-3x^2+2x-6$

Calculation:

The area of the first triangle is $x^3+4x^2+2x+8$

Find the factored form of this expression

Make group as shown below

  $\left(x^3+4x^2\right)+\left(2x+8\right)$

Now, factored out the GCF from each group

  $x^2\left(x+4\right)+2\left(x+4\right)$

Now, factored out the common term

  $\left(x^2+2\right)\left(x+4\right)$

Similarly, find the factored form of the second polynomial $x^3-3x^2+2x-6$

  $\begin{array}{l}\left(x^3-3x^2\right)+\left(2x-6\right)\\ =x^2\left(x-3\right)+2\left(x-3\right)\\ =\left(x^2+2\right)\left(x-3\right)\end{array}$

Now, the area of the rectangle is the product of length and width. And since the width of both the rectangles are equal.

Hence, the width will be the factor common in both of the factored form.

Thus, the width of the rectangles is $\left(x^2+2\right)$