# Homework Help: Are the set of all the polynomials of degree 2 a vector space?

1. Feb 2, 2012

### ironman1478

1. The problem statement, all variables and given/known data
Let P denote the set of all polynomials whose degree is exactly 2. Is P a vector space? Justify your answer.

2. Relevant equations
(the numbers next to the a's are substripts
P is defined as ---->A(0)+A(1)x+A(2)x^2

3. The attempt at a solution

I really dont know how to do this problem. i want to say that it isn't a vector space because it violates the property of having an additive inverse. as in, there is no value of x such that

F(x) + (F(-x)) = F(x)+(-F(x)) = 0 if we keep all of the values for A the same

A(0) + A(1)x + A(2)x^2 + A(0) + A(1)(-x) + A(2)(-x)^2 == 2A(0) + 2A(2)x^2 != 0

therefore, there is no additive inverse.
i probably did it wrong, but i dont know. all i know is that the book says that it isnt a vector space, but it doesnt give the reason.

2. Feb 2, 2012

### jbunniii

If p(x) is a 2nd degree polynomial, isn't -p(x) also a 2nd degree polynomial? And what happens when you add these two together?

3. Feb 2, 2012

### jbunniii

Can you think of two 2nd degree polynomials, p(x) and q(x), such that when you add them together, the resulting polynomial isn't 2nd degree?

4. Feb 2, 2012

### ironman1478

so because P(x) + (-(P(x)) = 0 and therefore, the answer is not a 2nd degree polynomial, then it cant be a vector space because it isnt closed under addition? if so, then i guess i just forgot to check the first property for a set to be a vector space and assumed it to be true.

5. Feb 2, 2012

### jbunniii

Yes, any vector space has to contain 0, and 0 isn't a 2nd degree polynomial.

Another example would be p(x) = x^2 + x + 1, and q(x) = -x^2. Then p(x) + q(x) = x + 1, which is 1st order.