The lengths of three sides of a quadrilateral are shown below: side 1: 3y2 + 2y − 6 side 2: 3y − 7 + 4y2 side 3: −8 + 5y2 + 4y the perimeter of the quadrilateral is 4y3 + 18y2 + 16y − 26. 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? part c: do the answers for part a and part b show that the polynomials are closed under addition and subtraction? justify your answer.

Answers: 
Part A: 12y² + 10y – 21
Part B: 4y³ + 6y² + 6y – 5 
Part C: See below.

Explanations: 
Part A: 
For this part, you add Sides 1, 2 and 3 together by combining like terms:
Side 1 = 3y² + 2y – 6
Side 2 = 4y² + 3y – 7
Side 3 = 5y² + 5y – 8
3y² + 2y – 6 + 4y² + 3y – 7 + 5y² + 5y – 8
Combine like terms: 
3y² + 4y² + 5y² + 2y + 3y + 5y – 6 – 7 – 8 
12y² + 10y – 21

Part B: 
You have the total perimeter and the sum of three of the sides, so you just need that fourth side value, which we can call d.
P = 4y³ + 18y² + 16y – 26
Sides 1, 2 & 3 = 12y² + 10y – 21 
Create an algebraic expression: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
– 12y²                                – 12y²
10y – 21 + d = 4y³ + 6y² + 16y – 26 
– 10y                               – 10y
– 21 + d = 4y³ + 6y² + 6y – 26 
+ 21                                + 21 
d = 4y³ + 6y² + 6y – 5 

Part C: 
If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y², y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms.

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