Question

the lengths of two sides of a triangle are shown.
side 1: 3x² - 4x - 1
side 2: 4x - x² + 5
the perimeter of the triangle is 5x³ - 2x² + 3x - 8.
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)

Explanation:

Step1: Add side 1 and side 2

$(3x^{2}-4x - 1)+(4x - x^{2}+5)$
$=3x^{2}-x^{2}-4x + 4x-1 + 5$
$=2x^{2}+4$

Step2: Find the third - side

Let the third - side be $s$. We know that perimeter $P$ of a triangle is $P=side1 + side2+s$. So $s=P-(side1 + side2)$.
Substitute $P = 5x^{3}-2x^{2}+3x - 8$ and $side1 + side2=2x^{2}+4$ into the formula.
$s=(5x^{3}-2x^{2}+3x - 8)-(2x^{2}+4)$
$=5x^{3}-2x^{2}-2x^{2}+3x - 8 - 4$
$=5x^{3}-4x^{2}+3x - 12$

Step3: Check closure property

In part A, we added two polynomials $3x^{2}-4x - 1$ and $4x - x^{2}+5$ and got $2x^{2}+4$, which is also a polynomial. In part B, we subtracted a polynomial $2x^{2}+4$ from a polynomial $5x^{3}-2x^{2}+3x - 8$ and got $5x^{3}-4x^{2}+3x - 12$, which is also a polynomial. So polynomials are closed under addition and subtraction.

Answer:

Part A: $2x^{2}+4$
Part B: $5x^{3}-4x^{2}+3x - 12$
Part C: Yes, because adding two polynomials in part A and subtracting polynomials in part B resulted in polynomials.