Linear algebra, polynomial problem Could someone help me with this question? Because I'm stuck and have no idea how to solve it & it's due tomorrow :(
Let $S$ be the following subset of the vector space $P_3$ of all real polynomials $p$ of degree at most 3:
$$S=\{p\in P_3\mid p(1)=0, p^\prime (1)=0\}$$
where $p^\prime$ is the derivative of $p$.
a) Determine whether $S$ is a subspace of $P_3$
b) determine whether the polynomial $q(x)= x-2x^2 +x^3$ is an element of S
Attempt:
I know that for the first part I need to proof that it's none empty, closed under addition and multiplication right?
will this give me full mark for the part a if I answer like this:
$(af+bg)(1)=af(1)+bg(1)=0+0=0$ and 
$(af+bg)′(1)=af′(1)+bg′(1)=0+0=0$ 
so therefore it's a subspace of $P_3$?
b) i got no idea...
Thank you very much!

(a) To show that $S \le P_3$, we just need to establish closure over addition, and scalar multiplication. We begin by letting $p, q \in S$, and fix $\theta \in \mathbb{R}$ (or any other field you're working with). Since $p$ and $q$ belong in $S$, we have $$p(0) = p'(0) = q(0) = q'(0) = 0.$$
Now, let $t = p \oplus q$. Then, $$t'(0) = p'(0) + q'(0) = 0,$$ so $t \in S$. Lastly, let $s= \theta \otimes p$. Then,
$$r'(0) = \theta p'(0) = 0$$
so that $r \in S$, as required. Hence, $S$ is closed over $\oplus$ and $\otimes$. Hence, $S \le P_3$.
(b) Let $p(x) = x^3 -2x^2 + x$. Then,
$$p(1) = 1 - 2 + 1 = 0,$$
and
$$p'(0) = 3 -4 + 1 = 0.$$
So, we must have $p \in S$.
