Originally Posted by

**Deveno** that is an entirely different, and much more specific question.

P2(R) has dimension 3 (one basis is {1,x,x^2}) so any OTHER basis for it must also have 3 elements. on the other hand, if you find 3 linearly independent elements of P2(R), you automatically have a basis.

if you want to find a basis for the subspace {p(x) : p(2) = 0}, first you have to determine ITs dimension. first of all, note that the only constant polynomial p(x) = c that satisfies p(2) = 0 is the 0-polynomial. that means our subspace can have dimension at MOST 2 (an entire dimension of P2(R) is not in our subspace). on the other hand, it has to have dimension at least 2, because x-2 and x^2 - 3x + 2 are both in it, and if:

a(x - 2) + b(x^2 - 3x + 2) = 0 <-- 0-polynomial, not the number

ax - 2a + bx^2 - 3bx + 2b = 0

bx^2 + (a-3b)x + (a+2b) = 0 so

b = 0

a-3b = 0

a+2b = 0, so a = b = 0, these two polynomials are linearly independent. so the dimension of your subspace is 2.

the first thing you must do, is determine which of your 5 polynomials are actually IN the subspace. you will be able to eliminate some of them. then, if you have more than 2 left, pick one. that will be the start of your basis. then pick another one, and test for linear independence. if 2 pass the linear independence test, you have a basis.