How to identify a basis for P2

Which of the following is a basis for P2 and why?

a. {x^2, x+1, x^2+x+1}
b. {x, 1, 0}
c. {x^2+x+1, x^2+x, x+1}
d. {x^2, x, 1, 0}

I know a basis exists when two conditions are met: linear independence and span, but I am unsure how to apply these ideas to the problem.

Thanks for looking.

The answer is choice c because it represents both a spanning set and a linearly independent set. For a deeper explanation read below.

P2 = the set of all polynomials of degree 2 or lower.

Since every such polynomial can be written as the linear combination ax^2 + bx + c(1) where a,b, and c are constants, we know that the set { x^2, x, 1 } spans P2.

Now for linear independence, suppose that
a(x^2) + b(x) + c(1) = the zero polynomial. The only way that can happen is if each of a, b, and c are zero. Thus the set {x^2, x, 1} is linearly independent.

Since the set {x^2, x, 1} is both a spanning set and linearly independent, it is a basis for P2.

Since this basis has 3 elements, P2 is three-dimensional (has dimension 3). Since all bases of finite-dimensional vector spaces have the same number of elements, we can rule out choice d which has 4 elements in it.

Choice b has 3 elements, but there is no way to combine x, 1, and 0 using only constants to obtain x^2 which is an element of P2. Therefore choice b does not span P2 (and is therefore not a basis). You can also show that it is not linearly independent because it contains 0 (any set containing 0 is not linearly independent).

Choice a is not a basis because it is not linearly independent. We can combine the first two elements to get the third element. Thus 1(x^2) + 1(x + 1) = 1(x^2 + x + 1) which can be rearranged to show that 1(x^2) + 1(x + 1) - 1(x^2 + x + 1) = 0. In other words we can combine these quantities using non-zero multipliers to get zero.

Thus the only remaining candidate is choice c. It has three elements and can be shown to be both a spanning set and a linearly independent set (though since it has the right number of elements, a theorem in linear algebra states that you only have to show one of the two properties and the other will follow).