Find a Basis for $W=\{p(x)\in V: p(1)=p'(1)=0\}$ 
Let $V=\mathbb{P_4}$ and $W=\{p(x)\in V: p(1)=p'(1)=0\}$. Assuming that $W$ is a subspace of $V$, find a basis for $W$ and thereby determine the dimension of $W$.

I think that $\dim(W)=3$ as there are two restrictions enforced upon $W$ $($ $p(1)=1$ and $p'(1)=0$$)$ and $\dim(\mathbb{P_4})=5$
However, I'm unsure of how to find a basis for $W$. I have tried to create components for my basis that satisfy the conditions of $W$, but is there a more concrete/routine way of finding my basis? I could not find solutions on this problem, so it is difficult to work backwards from a solution.
EDIT:
My solution is as follows.
Let $p(x)\in\mathbb{P_4}$, then $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$. Now \begin{align*}
p(1)&=a_0+a_1+a_2+a_3+a_4\\
p'(1)&=a_1+2a_2+3a_3+4a_4\\
\end{align*}
Hence $a_0=a_2+2a_3+3a_4$. Substituting this back into $p(x)$, we find $p(x)=a_1x+a_2(1+x^2)+a_3(2+x^3)+a_4(3+x^4)$.
Do the coefficients form my basis? If so, why?
 A: Hint
Let $p(x)\in \mathbb P_4$, i.e. $$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5.$$
Find condition on $a_i$ to have $p(1)=p'(1)=0$, and you'll find a basis (or at least a generating list), and you'll be able to find a basis as well.
A: $\def\env@matrix{\hskip -\arraycolsep
  \let\@ifnextchar\new@ifnextchar
  \array{*\c@MaxMatrixCols c}}$
You know two things, 
\begin{align}
a_0 + a_1 + a_2 + a_3 + a_4 = 0 \\ 
a_1 + 2a_2 + 3a_3 + 4a_4 = 0
\end{align}
It may be useful to look at these equations in terms of an augmented matrix, 
$$\begin{pmatrix}
  1 & 1 & 1 & 1 & 1 & 0\\
  0 & 1 & 2
& 3 & 4 & 0\end{pmatrix} $$
Where the last column denotes the zeroes after the equal signs. As we can see, there will be three free parameters in the solution set to these equations. Say, $a_2,a_3$ and $a_4$ vary freely. This hints that dim$W = 3$ as you suggested. 
From the second row/equation we get
\begin{align}
a_1 &= -\left(2a_2 + 3a_3 + 4a_4\right)
\end{align}
And back substituting into the first row/equation, we get 
\begin{align}
a_0 = a_2 + 2a_3 + 3a_4
\end{align}
as you derived. You were on the right track in your solution, substituting your equation for $a_0$ back into the polynomial and factorising in the manner you tried, though you were not using all the information available to you. 
If you use the equations for $a_0$ and $a_1$ above with your method, you will find that any polynomial $p \in \mathbb{P}_4$ (since you started with an arbitrary polynomial in $\mathbb{P}_4$ and went from there) satisfying the criteria to be a member of $W$ can be expressed as some linear combination of $p_1(t) = 1-2t + t^2$, $p_2(t) = 2-3t + t^3$ and $p_3(t) = 3 - 4t + t^4$. This implies that $W = \text{span}\{p_1, p_2, p_3\}$.
You may want to check that this set is independent for completeness. Since the set $\{p_1, p_2, p_3\}$ spans $W$, and is independent, $\{p_1, p_2, p_3\}$ forms a basis for $W$. The cardinality of this basis for $W$ is $3$, which implies $\text{dim}W = 3$.
A: Consider the linear map $F\colon\mathbb{P}_4\to\mathbb{R}^2$ defined by
$$
F(p)=\begin{bmatrix} p(1) \\ p'(1) \end{bmatrix}
$$
Then $W=\ker F$. The matrix of $F$ with respect to the standard basis $\{1,x,x^2,x^3,x^4\}$ is
$$
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 & 4
\end{bmatrix}
$$
which has the RREF
$$
\begin{bmatrix}
1 & 0 & -1 & -2 & -3 \\
0 & 1 & 2 & 3 & 4
\end{bmatrix}
$$
A basis for the null space of this matrix consists of three vectors, for instance
$$
\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}
\qquad
\begin{bmatrix}
2 \\
-3 \\
0 \\
1 \\
0
\end{bmatrix}
\qquad
\begin{bmatrix}
3 \\
-4 \\
0 \\
0 \\
1
\end{bmatrix}
$$
which correspond to the polynomials
$$
1-2x+x^2
\qquad
2-3x+x^3
\qquad
3-4x+x^4
$$
of which the vectors above are the coordinate vectors with respect to the standard basis, so they form a basis for $W$.
