if the wall is to be covered in wallpaper, how much wallpaper will be required?

Question

A wall with an area of (3x² + 4x + 1) square units has a rectangular window that has a width of (x + 2) units and a length of (x + 1) units.The wall also has a built-in shelving unit that occupies an area of (x² + 5x + 6) square units of wall space.

Answer 1

Hi William. This looks like one of those devious math problems meant to intimidate you with some crazy numbers. 😯

The problem as you gave it to us:

A wall with an area of (3x² + 4x + 1) square units has a rectangular window that has a width of (x + 2) units and a length of (x + 1) units.The wall also has a built-in shelving unit that occupies an area of (x² + 5x + 6) square units of wall space.

If the problem was written with ordinary numbers, it would read something like this:

A wall with an area of 200 square units has a rectangular window that has a width of 5 units and a length of 4 units.The wall also has a built-in shelving unit that occupies an area of 32 square units of wall space.

Easy peasy, when you look at it that way.

The process of solving the problem is the same regardless of how weird the numbers might look:

1. find the area of the window, with the measurements given for width (x+2) and length (x+1)
2. (x+2) * (x+1) = x² + 2x + x + 2 = (x² +3x + 2) square units for the window area
3. subtract the area of the window from the total area of the wall
4. (3x² + 4x + 1) - (x² +3x + 2)
5. 3x² + 4x + 1 - x² - 3x - 2 (remember, everything in the second equation flips signs)
6. 3x² - x² + 4x - 3x + 1 - 2 = 2x² + x - 1
7. subtract the area of the built in shelving from the remainder of the wall area
8. (2x² + x - 1) - (x² + 5x + 6)
9. 2x² + x - 1 - x² - 5x - 6 (flip the signs again on the second equation)
10. 2x² - x² + x - 5x - 1 - 6 = x² - 4x - 7 square units of wall paper.

Does that make more sense?