The lengths of two sides of a triangle are shown below: 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? (4 points) Part B: What is the length of the third side of the triangle? (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)
@Hero can you help?
@sleepyjess @sammixboo
@ganeshie8
which part are you stuck at ?
I think i have a and b but i want to make sure im right
good, what do you have for a and b ?
for a i have 2x^2+4 and part b i have 5x^3-4x^2+3x-4
part a is right part b is wrong, but you're very close !
ok that's what I thought but what did I do wrong? can you explain
we're given two sides: Side 1: \(3x^2 - 4x - 1\) Side 2: \(4x - x^2 + 5\) For part \(a\) : Sum of the two sides can be obtained by adding both the sides : Side1 + Side2 = \(3x^2 - 4x - 1\) + \(4x - x^2 + 5\) = \(\color{blue}{2x^2 + 4}\)
For part \(b\) : Since the perimeter is given as \( 5x^3 - 2x^2 + 3x - 8.\), subtracting the sum of two sides from part \(a\) gives the third side : \[ 5x^3 - 2x^2 + 3x - 8- (\color{blue}{2x^2+4}) = 5x^3-4x^2+3x-12\]
oh ok I see where I got it wrong
So are we convinced adding/subtracting two polynomials gives another polynomial ?
no
wait yes sorry. I switched the question around
why not ? In part \(a\), we have added two polynomials (\(3x^2 - 4x - 1,~~4x - x^2 + 5\)) and got another polynomial : \(\color{blue}{2x^2 + 4}\). This kinda shows that adding two polynomials gives another polynomial, yes ?
Ohk :)
so is that what i put that for part c? or is it something different?
Yes and add something like this : Similarly part \(b\) convinces us that "subtracting" two polynomials gives us another polynomial. So can we say part \(a\) and part \(b\) show us that the polynomials are closed under addition and subtraction.
ok thanks soooooooo much!!!!!!!!!! and do you think you can help me with one more? its my last question @ganeshie8
Okay please ask in a separate thread :)
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