Finding image and kernel of this set 
Consider $\mathbb{R_4}[x]$ the space of polynomial functions with degree less or equal than 4. Consider the linear transformation $T: \mathbb{R_4}[x] \rightarrow \mathbb{R_4}[x]$ defined as $T(p(x)) = p(x+1) - p(x)$. Calculate $N(T)$ and $I(T)$

I have absolutely no idea how can I start this resolution... 
My professor said to us to not use matricial representation so please do not suggest that (I know it becomes easier)... Anyway, can someone give an hit to start the problem or just what I need to do? 
Thanks! 
 A: Kernel :
$p(x+1)=p(x)$ implies that your function is periodic with period 1. This is only true of constant polynomials, so the kernel is $\mathbb R_0[X]$
Image :
Consider the image of each basis element :
$$T(X^n)=(n-1)X^{n-1}+Q(X)$$
for $n\neq 0$ and with $Q$ a polynomial of degree $<n-1$. This means that $\deg (T(X^n))=n-1$. Therefore, the image of the basis is a family of polynomials of distinct orders 3,2,1,0. They generate $\mathbb R_3[X]$.
TL;DR : Ker = $\mathbb R_0[X]$, Im = $\mathbb R_3[X]$
A: Clearly the line
$$p(x+1) - p(x)$$
Can only be $0$ if $p(x+1) = p(x)$. Every polynomial is a well defined function, meaning it is continuous everywhere, but what set of functions easily answer this?
Hint
$$a_4(x - 1)^4 + a_3(x - 1)^3 + a_2(x - 1)^2 + a_1(x - 1) + a_0  - b_4x^4 + b_3x^3 + b_2x^2 + b_1x + b_0 = 0$$
What conditions need to be met? It seems that there's only a single polynomial basis. Now for imagespace, does the transformation actually change the degree of the polynomial?
Note: Set of all constants functions?
Note: The imagespace is also the set of all polynomials with degree less than 4. 
