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.
