Let $U = { p\in \mathbb{P_4}(\mathbb{R}): p''(6) = 0}$. Find a basis for U Let $U = \{ p\in \mathbb{P_4}(\mathbb{R}): p''(6) = 0\}$. Find a basis for U
I was given a hint of how to figure this out, but I can't seem to get how things work together: 
For polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$, consider the condition when $f”(6)=0$. Then you will get a linear equation about $a,b,c,d,e$. Find a basis of the solution space of this linear equation. Then substitute it into $f(x)=ax^4+bx^3+cx^2+dx+e$, you will get a basis of U
Attempt: I don't fully understand what solution space they are talking about. When considering $f''(6) = 0$, I thought of this: $$f''(x) = 12ax^2+6bx+2c$$
So if I'm supposed to consider $f''(6) = 0$ this means I should be able to find another basis element by the expression: $$\frac{12ax^2+6bx+2c}{(x-6)}$$.
Unfortunately this doesn't work out for me......
Also, what is the "division" command in latex?
 A: Start with $p(x)=ax^4+bx^3+cx^2+dx+e$. As you observed, you do get
$$p''(x)=12ax^2 + 6bx+2c$$
Now set $x=6$ and use the fact that $p''(6)=0$. You get 
$$432a + 36b+2c=0$$
Now, you need to try to express $c$ in terms of $a$ and $b$ (you can do this conveniently by first setting $a=1$ and $b=0$, and then finding $c$, and then setting $a=0$ and $b=1$, and then finding $c$), which you can then plug back into your original polynomial with $d=e=0$ to get two basis elements.
Finally, don't forget the last two basis elements:
If $p(x) = 1$ or $p(x) = x$,  then $p''(x)=0$ for all $x$, so $p''(6)$ is trivially satisfied.

Using this method gives the basis:
$$\{1,x,x^3-18x^2, x^4-216x^2\}$$
(This basis can be shown to be equivalent to the basis in the other answer; it might be helpful for you to try to justify this.)
A: The vectors $(x-6)^k$ $\>(0\leq k\leq4)$ form a basis of $P_4$. Therefore the general element $p(x)$ of $P_4$ can be written as
$$p(x)=\sum_{k=0}^4 a_k(x-6)^k,\qquad a_k\in{\mathbb R}\quad (0\leq k\leq4)\ .$$
One computes
$$p''(6)=2a_2\ .$$
It follows that $p''(6)=0$ is equivalent with $a_2=0$. A basis of $U$ is therefore given by
$$\bigl(1, \ x-6, \ (x-6)^3, \ (x-6)^4\bigr)$$
A: \begin{align}p''(6) &= 0 \text { and } p \in P_4 \to  \\
p''(x) &= 12a(x-6)^2 + 6b(x-6)\\
p'(x) &= 4(x-6)^3+3b(x-6)^2+c\\
p(x) &= a(x-6)^4+b(x-6)^3+c(x-6)+d\end{align}
therefore the subspace $U = \{ p\in \mathbb{P_4}(\mathbb{R}): p''(6) = 0\}$ is of dimension four and a basis is $\{(x-6)^4, (x-6)^3, (x-6), 1\}.$
