Question

WORTH 90 POINTS!

The lengths of three sides of a quadrilateral are shown below:

Side 1: 4y + 2y2 − 3

Side 2: −4 + 2y2 + 2y

Side 3: 4y2 − 3 + 2y

The perimeter of the quadrilateral is 22y3 + 10y2 + 10y − 17.

Part A: What is the total length of sides 1, 2, and 3 of the quadrilateral? (4 points)

Part B: What is the length of the fourth side of the quadrilateral? (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)

Answer 1

A. total length of side 1-3 = side 1 + side 2 + side 3

Side 1: 4y + 2y2 − 3

Side 2: −4 + 2y2 + 2y

Side 3: 4y2 − 3 + 2y

so total = 8y^2 + 8y - 10

B. perimeter = sum of side 1-4

so length of side 4 = perimeter - total of side 1-3

= 22y3 + 10y2 + 10y − 17 - (8y^2 + 8y - 10)

= 22y^3 + 2y^2 + 2y - 7

C. they are close for addition and subtraction. Part A and B show that.

Answer 2

A. Just add the given sides.

8y^2+8y-10 is the answer.

B. The perimeter of any shape is all the sides added together. The perimeter given is the perimeter of the quadrilateral, and we are to find the other side. We are given values of three sides. We can find the value of the other side by subtracting the sum of those three sides from the total perimeter:

(22y^3+10y^2+10y-17)-[(4y+2y^2-3)+(-4+2y^2+2y)+(4y^2−3+2y)]

When simplified, you get 22y^3+2y^2+2y-7.

C. The polynomials are closed under addition and subtraction as the result is another polynomial. It can be justified by parts A and B.

Hope this helps.