How to prove that a given set of functions forms a linear subspace of a larger space? I need to prove that these $S$s are Linear subspaces of $V$. 
I tried:
1- I was thinking: I can choose any scalar number who belongs to real numbers, so, if I to multiply $p(x)$ by a scalar, I have a $p(x)$ that belongs to Real Numbers. The addition is ok. The fact of p has at least one real root it means that this graph must be real? 
2- I was thinking about a function likes $(x,0)$, where I can say: given $u=(x_1, y_1)$ and $v=(x_2,y_2)$, we have $(x_1+x_2, y_1+y_2)= (x_1+x_2, 0)$, and $\alpha(x_1,0)= (\alpha  x_1 ,0 \alpha)$ But, I have no idea about another cases.
If I get the same integral without the square, how can I deal with it?
3- I know that we have $$ p=ax^3 + bx^2 + cx +d $$ where $b$ is null.
Given $$a_1x^3 + b_1x^2 + c_1x +d_1 $$ and $$a_2x^3 + b_2x^2 + c_2x +d_2 $$
where the x of c is null. I have to prove the addition and the scalar multiplication.
How can I  represent it using $u$ and $v$, given $v=(x_1, y_1)$ and $u=(x_2,y_2)$? 
About the addition, I can say :
$$ p=(a_1+a_2)x^3 +(b_1 +b_2)x^2 + (c-1+c_2)x + (d_1+d_2) $$
The $c$ need be $0$?
If I to multiply $p$ by $\alpha$, I know that it works.


$\textbf{Some hints}$:
$0$.In $1$ $S$ isn't a vector subspace, in $2,3$ is.
$1$.You know that odd degree polynomial always has at least one real root, so $p(x)=-x^5$ and $q(x)=x^5+x^2+1$ have  at least one real root. What about $p(x)+q(x)$?
$2$.First note that for all $f,g \in S \Leftrightarrow f(x)=g(x)=0$ for $x \in [0,1]$ (because $f,g$ are countinous). What about $f+g$ and $\alpha f$ where $\alpha \in \mathbb{R}$?
$3$.Let $f,g \in S$, so the coefficient in the $x$ term is null. Prove that $x$ coefficient in $f+g$ and $\alpha f$ is also null, so $f+g \in S$ and $\alpha f \in S$.
