A Subspace of the Degree $3$ Polynomials Space for which $P(5)=0$ The problem is the following:

Given the space of the polynomials $V=P(x)$ of degree $3$ prove the following:
  
  
*
  
*The set $U$ which is defined as the set of the elements for which $P(5)=0$ is a subspace of $V$.
  
*Find a basis for $U$.
  
*If it was asked that $P(x)$ has $2$ also as a root then show that the new subset $W$, which is defined as $W=\{ p(x):p(x)=(x-5)(x-2)(b_0+b_2x)\}$ is also a subspace of $V$ and find a basis for it.
  

So, I do face difficulties in the first and the last one. I wrote down the elements of $V$:
\begin{equation}
V= \{ P(x)=a_1x^3+b_1x^2+c_1x+d_1, (a_1,b_1,c_1,d_1) \in \mathbb{R} \}
\end{equation}
Then I took the elements of $U$ to be of the form:
\begin{equation}
V= \{ p(x) \in \mathbb{R}:p(x)=(x-5)(a_2x^2+b_2x+c_2), (a_2,b_2,c_2) \in \mathbb{R} \}
\end{equation}
since $5$ is a root of $P(x)$. But I do not know how to proceed. I mean I know that I have to prove that $U$ has the properties of a subspace, but should I find the connection between the coefficients $a_i, b_i, c_i$? And how to use that relation?
Any ideas? 
Thank you!
 A: There is no relation between the coefficients you use: these $a_1,b_1,c_1,d_1$ and $a_2,b_2,c_2$ can be arbitrary.
Hint:
One way to prove that $U$ is a subspace is to find a linear map with kernel exactly $U$.

 $V\to\Bbb R,\ f\mapsto f(5)$

You can also use it for 3.
A: $1$. Let $\mathcal{P}_1(x)=\sum a_ix^i,\mathcal{P}_2(x)=\sum b_ix^i\in U$ then $[\mathcal{P}_1 + \mathcal{P}_2](5)=\sum (a_i+b_i)5^i=\sum a_i5^i+\sum b_i5^i=0+0=\mathcal{P}_1(5)+\mathcal{P}_2(5)$.
Can you prove the second property from here?
2.
By definition $a_0+5a_1+5^2a_2+5^3a_3=0\Rightarrow a_0=-5a_1-25a_2-125a_3$.


*

*Claim that the set $T$={$-5a_1-25a_2-125a_3,xa_1,x^2a_2,x^3a_3$} is a basis for $U$.
It's easy to show that the set is L.I. since it has only 3 free variables it's dimension is 3 ,now we have to show that $dim(U)=3$.Suppose that $dim(U)=4$ then it should contain every polynomial of degree at least 4,since $\mathcal{P}(x)=x-2\notin U$ it follows that $dim(U)<4$ and $spam(T)\subseteq U$
 it follows that $dim(U)=3$ hence $T$ is a basis.
