Yun was trying to factor

Question

. He found that the greatest common factor of these terms was

and made an area model:
An area model for a rectangle with a height of seven x and an unknown width. The rectangle is broken into two rectangles to isolate each term in the width. The first rectangle has a height of seven x and an area of seven x squared. The second rectangle has a height of seven x and an area of negative fourteen x.




What is the width of Yun's area model?

Answer 1

To find the width of Yun's area model, we start by looking at the area of each rectangle based on the provided information.

1. The first rectangle has a height of \( 7x \) and an area of \( 7x^2 \). We can calculate the width of this rectangle using the formula for area: \[ \text{Area} = \text{Height} \times \text{Width} \] So, for this rectangle: \[ 7x^2 = 7x \times \text{Width}_1 \] To find \( \text{Width}_1 \): \[ \text{Width}_1 = \frac{7x^2}{7x} = x \]

2. The second rectangle has a height of \( 7x \) and an area of \( -14x \). We can similarly calculate the width for this rectangle: \[ -14x = 7x \times \text{Width}_2 \] To find \( \text{Width}_2 \): \[ \text{Width}_2 = \frac{-14x}{7x} = -2 \]

3. Now, the total width of the rectangle, which combines the widths of both rectangles, is given by: \[ \text{Total Width} = \text{Width}_1 + \text{Width}_2 = x + (-2) = x - 2 \]

Thus, the width of Yun's area model is \( x - 2 \).