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The lengths of three sides of a quadrilateral are shown below: Side 1: 1y2 + 3y − 6 Side 2: 4y − 7 + 2y2 Side 3: 3y2 − 8 + 5y The perimeter of the quadrilateral is 8y3 − 2y2 + 4y − 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? (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)

This is alegbra 1 9th grade i just need some help pleaseeeeeee

1 Answer

6 votes

Answer:

Part A : Total length of Given 3 side =
6y^2+12y-21

Part B : Length of side 4 =
8y^3-8y^2-8y-5

Part C : Yes, Part A & Part B shows that Polynomials are closed under addition and subtraction

Explanation:

Given: Sides of a quadrilateral, Side 1 =
y^2+3y-6 , Side 2 =
2y^2+4y-7,

Side 3 =
3y^2+5y-8.

Perimeter of Quadrilateral =
8y^3-2y^2+4y-26

To find: [A] Total length of given 3 sides.

[B] Length of Side 4.

[C] Do part A & B show that the polynomials are closed

under addition and subtraction?

Part A -

Total length of Given 3 side = Side 1 + Side 2 + Side 3

=
y^2+3y-6+2y^2+4y-7+3y^2+5y-8

=
y^2+2y^2+3y^2+3y+4y+5y-6-7-8

=
(1+2+3)y^2+(3+4+5)y+(-6-7-8)

=
6y^2+12y+(-21)

=
6y^2+12y-21

Part B -

Length of side 4 = perimeter - total length of 3 sides

=
8y^3-2y^2+4y-26-(6y^2+12y-21)

=
8y^3-2y^2+4y-26-6y^2-12y+21

=
8y^3-2y^2-6y^2+4y-12y-26+21

=
8y^3+(-2-6)y^2+(4-12)y-26+21

=
8y^3+(-8)y^2+(-8)y-5

=
8y^3-8y^2-8y-5

Part C -

Yes, Part A & Part B shows that Polynomials are closed under addition and subtraction because after addition and subtraction result is also a polynomial.

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