Writing a polynomial for this box thing

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To write a polynomial representing the area of the shaded region, one must first calculate the areas of both the outside and inside boxes. The area of the shaded region can be found by subtracting the area of the inside box from the area of the outside box. A polynomial is defined as a function involving powers of the variable x, and it is important to distinguish between area and perimeter, as the area is calculated using length times width rather than the perimeter formula. Clarification on mathematical terminology is crucial for understanding and solving these types of problems. Understanding these concepts will aid in accurately modeling the area with a polynomial.
ceres_sun
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Could someone guide me through this? *math dummy*

Write a polynomial in standard form that models or represents the area of the shaded region.

http://www.sfu.ca/~tca19/thing.GIF

Thanks!
 
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What is the area of the outside box?

What is the area of the inside box?

How would you get the area of the shaded region?
 
|2(x-4)+2(x+3)| - |2(x+2) + 2(x+1)|

I'm actually more confused as to what a polynomial is >_> *cough*
 
A polynomial is a function that only involves powers of the variable x.

However, your real problem is that you are confusing "area" with "perimeter". The area of a rectangle is "length times width", lw, not 2l+ 2w. Also, since all the numbers involved are positive, there is no need for absolute value.
 
LOL thank you so much.

Yes, my main problem with math is that I don't know what most of the terms mean and/or confuse them with other terms...-_-

I need a math dictionary.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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