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The lengths of two sides of a triangle are shown.

Side 1: 8x^2 − 5x − 2

Side 2: 7x − x2 + 3

The perimeter of the triangle is 4x^3 − 3x^2 + 2x − 6.

Part A: What is the total length of the two sides, 1 and 2, of the triangle? Show your work. (4 points)

Part B: What is the length of the third side of the triangle? Show your work. (4 points)

Part C: Do the answers for Part A and Part B show that the polynomials are closed under addition and subtraction? Justify your answer. (2 points)

User Edward Newell
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1 Answer

29 votes
29 votes

Answer:

Explanation:

Length 1 = 8x^2 - 5x - 2

Length 2 = 7x - x^2 + 3

Part A

Length of the two sides = 8x^2 - 5x - 2 + 7x - x^2 + 3

Length of the two sides = 8x^2 - x^2 - 5x + 7x - 2 + 3

Length of the two sides = 7x^2 + 3x + 1

Part B

The perimeter is 4x^3 - 3x^x + 2x - 6

The Third side must be Perimeter - the sum of the two sides.

The third side must be 4x^3 - 3x^x + 2x - 6 - 7x^2 - 3x - 1

The third side = 4x^3 - 4x^x -7x^2 - x - 7

Part C

The problem is now does this show closure. I would say it does not. I don't think 4x^x is in the quadratic family. Let's take a simpler example of closure.

Suppose we say 4 + 5 = 9 and we say further that we are working with all the integers. 9 is an integer and so are 4 and 5. So the integers are closed by addition.

4 - 5 = - 1 The answer (-1) is still an integer, so the example still shows closure.

But the question has things in it that show closure only if the set of numbers is the real numbers and even then I'm not certain. We have to ask what x^x is.

User Brian Lyttle
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3.2k points