Determine whether S is a subspace of P3. Vector space of all real polynomials. 
ATTEMPT:
Have given a small attempt just really confused on how to approach.
So I got the general equation of $p(x)= a + bx +cx^2 +dx^3$.
So we find the derivative? and find the values of $a,b,c,d$? 
How do I find the polynomials? like the 2nd equation?
My working is just all over the place and then I know what to do next with the addition and scalar multiplication steps. Just not sure on preliminary steps. 

Any element of the vector space $P_3$ is of the form,
$p(x)= a + bx +cx^2 +dx^3$
Substituting x = 1, we get,
$p(1)= a + b.1 + c.1^2 +d.1^3$
$\implies a + b + c + d = 0$
For p'(1), we have after differentiating,
$p'(x)= b + 2cx +3dx^2$
$p'(1)= b + 2c.1 +3d.1^2 = 0 $
$\implies b + 2c +3d = 0$
So, you know every element of S follows these two relations. Now assume two vectors $u, v$ of S with real coefficients
$u = a_1 + b_1 x + c_1 x^2 + d_1 x^3$ and $v = a_2 + b_2 x + c_2 x^2 + d_2 x^3$
and try to prove whether or not it is a subspace by testing whether S is closed under vector addition and scalar multiplication, i.e. For any vector $r$ in $P_3$, such that
$r=v+u={(a_1+a_2)} + {(b_1+b_2)}x + {(c_1+c_2)}x^2 + {(d_1+d_2)}x^3$
We can find constants $a_3, b_3, c_3, d_3$ in $R$ such that $a_3 = a_1 +a_2$ and so on. That means,
$r=a_3+b_3x+c_3x^2+d_3x^3$
Since we know that both $a_1+b_1+c_1+d_1=0$ and $a_2+b_2+c_2+d_2=0$, we can infer that $a_3+b_3+c_3+d=0$ and similarly that $b_3+2c_3+3d_3=0$. This means that the vector $r$ lies in the subset $S$, and hence $S$ is closed under vector addition.
Using similar arguments, we see that S is also closed under scalar multiplication, and so, $S$ is a subspace of $P_3$.
For part (B): you need to verify whether the two coefficient conditions are satisfied for $q(x)$, that is,
$a+b+c+d = 0 + 1 + (-2) + 1 = 0$
and 
$b+2c+3d = 1 + 2(-2)+3(1) = 0$
which implies $q(x)$ is an element of $S$.
