Subspace, Direct Sum, Polynomials, Basis 
Let $U = \{p \in \mathcal{P}_4(\mathbb{F}) \;\colon\; p''(6) =  0\}$.
  a. Find a basis for $U$.
  b. Extend the basis in part (a) to a basis for $\mathcal{P}_4(\mathbb{F})$.
  c. Find a subspace $W$ of $\mathcal{P}_4(\mathbb{F})$ such that
  $\mathcal{P}_4(\mathbb{F}) = U \oplus W$.

If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $\mathcal{P}_4(\mathbb{F})$.
Please let me know if I am correct.
Moreover I am stuck with the part (c).
 A: If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
\begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\\
p''(x)&=2a_2+6a_3x+12a_4x^2
\end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by 
$$
\{1,x,-18x^2+x^3,-216x^2+x^4\}
$$
Now you have just to find a polynomial $q$ such that $q\notin U$, for instance $q(x)=x^2$.
Add this to the above basis; of course, you can also take $W=\operatorname{Span}(x^2)$.
So you're correct. I just found a different basis, with a more systematic approach.
A: Let us assume $\mathbb F=\mathbb R$. (Or $\mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)
You could use the fact that very polynomial in $\mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
\begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
\end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have 
$$U=\{p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3\in\mathbb F\}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.

The only difference is if $\mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=\mathcal P_4$.
