Find the basis for a kernel and the dimension of the image for a linear map 1. The problem statement, all variables and given/known data
Let $n>1\in\, \mathbb{N}$. A map $A:\mathbb{R}_{n}[x]\to\mathbb{R}_{n}[x]$ is given with the rule $(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)$
a)Prove that this map is linear
b)Find some basis of the kernel
c)Find the dimension of the image
2. Relevant equations
$\mathbb{R}_{n}[x]$
is defined as the set of all polynomial with real coeficient that have the power less or equal to n
$kerA=\{x;Ax=0\}$
$imA=\{Ax,x\in\mathbb{R}_{n}[x]\}$
$p(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^2+a_{1}x^1+a_{0}x^0$
3. The attempt at a solution
a)Proof that this map is linear
$p,q\in A \\
(A(p+q))(x)=(x^n+1)(p+q)(1)+(p+q)^{'''}(x)=((x^n+1)(p)(1)+(p)^{'''}(x)+(x^n+1)(q)(1)+(q)^{'''}(x)=Ap(x)+Aq(x)
\\
\text{    }
\\
A(\theta p)(x)=\theta((x^n+1)(p)(1)+(p)^{'''}(x))=\theta(Ap)(x)
$
c)
I started doing this by writing out the possible polynomial
$(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)\\
p(1)=a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\ldots+a_{2}+a_{1}+a_{0}\\
p^{'''}(x)=6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
$
now I placed all of this together
$
(Ap)(x)=(x^n+1)\displaystyle\sum_{i=0}^{n}a_{i}+6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
$
Then I paired all of the same coeficients together and got
$$
a_{n}(x^{n}+1+6\binom{n}{3}*x^{n-3}) \\
a_{n-1}(x^{n}+1+6\binom{n-1}{3}*x^{n-4}) \\
a_{n-2}(x^{n}+1+6\binom{n-2}{3}*x^{n-5}) \\
\vdots\\
a_{3}(x^n+7) \\
a_{2}(x^n+1) \\
a_{1}(x^n+1) \\
a_{0}(x^n+1) \\
$$
Here I noticed that the bottom 3 functions are linearly dependent, which means that If I want to find the basis  or dimensions I should take $a_{0}\, and\, a_{1}$ out. Then I also noticed that all of the above  (a[SUB]n[/SUB] to a[SUB]3[/SUB]) are also linearly dependent on a[SUB]2[/SUB] so I subtracted them and got
$$
a_{n}(x^{n-3}) \\
a_{n-1}(x^{n-4}) \\
a_{n-2}(x^{n-5}) \\
\vdots\\
a_{3}(6) \\
a_{2}(x^n+1) \\
$$
as my basis for the image of A therefore the dimension(imA)=n-1
This is as far as I have gotten. 
I don't know if this is correct but I really don't know how to continue
I would really appreciate if someone could show me how to solve b) and possibly check If what I did is even correct.
Thank you
 A: Your solution so far seems correct.
Using the dimension theorem - as commented - you will get the dimension of the kernel is 2.
However, you found a basis for the image, which was not asked and not for the kernel, which was asked. 
Well, you can finish from here, just pick any 2 indepent vectors from the kernel.

a) For a fixed polynomial $q$ and number $x_0$, the functions $p\mapsto p(x_0)$ and also $p\mapsto p(x_0)\cdot q$ are linear. The differential $D:p\mapsto p'$ is also linear, hence so is $D\circ D\circ D=p\mapsto p'''$, and thus any linear combinations of them are linear.
b) Let's find the kernel. What does $Ap=0$ mean? 
Observe that $\deg p'<n$, so  $\ p(1)\cdot(x^n+1)\ +\ p''' \ =\  0\ $ implies $p(1)=0$, else $x^n$ would have nonzero coefficient.
This in turn implies $p'''=0$, too.
So, the kernel consists of at most quadratic polynomials ($p'''=0$) that vanish at $1\ $ ($p(1)=0$).
The polynomials which vanish at $1$ are exactly the multiples of $(x-1)$.
So, a basis for the kernel is:
$( x-1, \ (x-1)\cdot x)$
c) Use the dimension theorem to obtain the $n-1$, in accordance with your basis $\big(6, x^n+1,\ x, x^2, x^3,\dots, x^{n-3} \big)$.
