Finding a base and dimension to a polynomial space under conditions I need to find a base and dimension for the polynomial space $V$ while the factor of $x$ is 58: $$V=\left \{  \right.p(x)\in P_{4}[x]: p(1)=p''(1)=0\left.  \right \}$$
What I did is to  create a general polynome: $p(x)=e+dx+cx^2+bx^3+ax^4$
Then I calculated:
$$p(1)=e+d+c+b+a=0$$
$$p''(1)=2c+6b+12a=0$$
I've expressed $p(x)$ as bellow: $$p(x)=(2/3c+e+58)x^4+(-3/5c-2e-116)x^3+cx^2+58x+e$$
Here I'm stuck because I don't know how to continue.
Help will be more than appreciated!
 A: Let's say that $p(x)=\sum_{k=0}^4 a_kx^k$.
The conditions are:
$$p(1)=\sum_{k=0}^4 a_k=0$$
$$p''(1)=\sum_{k=2}^4 k(k-1)a_k=0$$
So $V$ is the kernel of the linear map
$$\begin{pmatrix}1&1&1&1&1\\12&6&2&0&0\end{pmatrix}\begin{pmatrix}a_4\\a_3\\a_2\\a_1\\a_0\end{pmatrix}$$
Since the rank of this map is $2$, $\dim V=5-2=3$.
Now you must find three l.i. polynomials that meet the conditions.
A: Solution:
Let's write a general polynomial: $p(x)=ax^4+bx^3+cx^2+dx+e$
Then: $p''(x)=12ax^2+6bx+2c$
Under the condition $p''(1)=p(1)=0$: $$a+b+c+d+e=0$$ $$12a+6b+c=0$$
Let's express $a$ and $b$ using $c,d=58,e$: $$a=2/3c+58+e$$ $$b=-5/3c-116-2e$$
$$\Rightarrow p(x)=(2/3c+58+e)x^4+(-5/3c-116-2e)x^3+cx^2+58x+e$$
$$\Rightarrow p(x)=sp\left \{ 58(x^4-2x^3+x)+c(2/3x^4-5/3x^3+x^2)+e(x^4-2x^3+1) \right \}$$
We'll check that our polynomials are l.i.: $$\begin{pmatrix}
 1 & -2 & 0 & 1 & 0 \\ 
2/3 & -5/3 & 1 & 0 & 0 \\ 
 1 & -2 & 0 & 0 & 1
\end{pmatrix} \xrightarrow[r2\rightarrow r2-2/3r1]{r3\rightarrow r3-r1}\begin{pmatrix}
1 & -2 & 0 & 1 & 0\\ 
0 & -1/3 & 1 & -2/3 & 0\\ 
0 & 0 & 0 & -1 & 1
\end{pmatrix}
$$ 
$\Rightarrow$The polynomials are l.i.
$\Rightarrow$A base for V while the coefficient of $x$ is 58: $\left \{ (58x^4-116x^3+58x),(29x^3-87x^2+58x),(-58x+58) \right \}$
