Well i get 1. , i've done it by choosing two general equations as F(x) and G(x) which are both in space S
then using those proved axiom A1 and S1 , (f+g)(1)=f(1)+g(1)=0 and same way for the derivative, but i'm not sure if it's right or not
and i can't seem to figure out what this represents geometrically :S
Since P(1) = 0, it means P has a root at x=1. However, since P'(1) = 0 too, it means that the root at x=1 is a double root.
So S is the set of all polynomials (of degree at most three) satisfying P(1)=0 and P'(1)=0. For a quadratic and cubic, this means it has a double root at x=1.
By A1 and S1, I presume you mean the addition and scalar properties respectively?
I'll give you the 'English translation' to make things a bit easier for you. Seems like you're having difficulty understand what the question is asking. You already know that to prove that S is a subspace of P_3, you have to prove 3 properties.
Addition: If you add any two such polynomials (satisfying the condition that P(1)=P'(1)=0), you should also have a polynomial that still satisfies that condition.
Scalar: If you multiply a polynomial in S by a scalar, it still satisfies the above condition.
Zero Vector: The zero vector (in this case the zero polynomial) is in S.
To prove S is a subspace of P_3, you need to show that S satisfies all these three properties. Give it a try now!