Answer:
a. The possible dimensions of the total area are;
The width of the total area is (2·x + 8) inches
The length of the total area is (2·x + 10) inches
b. The dimensions of the photos are as follows;
The length of the photos are 10 inches
The width of the photos are 8 inches
Explanation:
a. Given that the area, A = 4·x² + 36·x + 80
We get;
A = 4 × (x² + 9·x + 20) = 4 × (x + 4) × (x + 5) = (2·x + 8)·(2·x + 10)
Therefore, the possible dimensions of the total area (photo + mat) are;
The width of the total area (photo + mat) = (2·x + 8) in.
The length of the total area (photo + mat) = (2·x + 10) in.
b. The dimensions of the photos alone are shorter than the dimensions of the photo and mat combined by 2·x each
Therefore, we have the dimensions of the photos are as follows;
The length of the photo = (2·x + 10) in. - 2·x in. = 10 in.
The width of the photos = (2·x + 8) in. - 2·x in. = 8 in.