$E$ is a certain subspace of $\mathbb{R}[x]$. Is the set $\{x − 2, (x − 2)^2, (x − 2)^3\}$ a basis of $E$? The question goals

Recall that we use the symbol $\mathbb R[x]$ to mean the real vector space of all polynomials in $x$ with real coefficients. Show that the set
  $$
E=\{p(x)\in \mathbb{R}[x]:\, p(2)=0 \, \, \text{and} \, \, \deg(p(x))\le 3\}
$$
is a subspace of $\mathbb{R}[x]$. Is the set $\{x − 2, (x − 2)^2, (x − 2)^3\}$ a basis of $E$? 

I'm a bit confused. I had a look at my notes I think I might be able to use 6 of the 10 "vector space axioms". But that's just because it's a similar question. I'm not too sure with this content. I was going to go look for a linear combination. 
 A: The first part has already been answered in the comment by uer84413.
As for the second part: calculate the polynomials $(x-2)^2$ and $(x-2)^3$. Then it should be easy for you to prove that $x-2, (x-2)^2$ and $(x-2)^3$ are linearly independent. 
A: You only need these three conditions to check if it is a subspace-(first one is not necessary but rather fun)


*

*$0\in E$, clearly.

*Closure under addition- Let $a(x),b(x) \in E$, then $a(2)+b(2)=0+0=0$ and $\deg(a+b) \le \max\{\deg(a),\deg(b)\} \le 3$, hence satisfied.

*Closure under scalar multiplication- If $a\in \Bbb{R}$ and $p(x)\in E$, does $g(x)a.p(x)\in E$? Clearly $g(2)=0$, check the other condition too.


So it is a subspace.
To check the set $S=\{x − 2, (x − 2)^2, (x − 2)^3\}$ a basis try proving linear independence i.e. $\sum_{i=1}^3 a_i(x-2)^i=0 \iff a_i=0\ \forall\ i=1,2,3$ by expanding and comparing coefficients and to prove that every element of $E$ lies in span of $S$, let $y=x-2$ and prove $\{y,y^2,y^3\}$ spans $E'=\{f'(x)\in \Bbb{R}[x] : f'(0)=0 \text{and}\ \deg\ f'(x) \le 3\}$ as $E\cong E'$
A: I'm going to guide you. $E$ is a nonempty subset of $\mathbb{R}[x]$ ($0\in E$). So to prove that $E$ is a subspace of $\mathbb{R}$, why not simply use the following theorem?

Subspace Criterion
Let $S$ be a subset of the real vector space $V$. Then $S$ is a subspace of $V$ if and only if:
$(i)\,\,0\in S$
$(ii)\,\,\forall v_1,v_2\in S,\,v_1+v_2\in S$
$(iii)\forall v\in S,\,\forall \alpha\in\mathbb{R},\,\alpha v\in S$

Now $\mathcal{B}=\{(x-2),\,(x-2)^2,\,(x-2)^3\}$ is a basis of $E$, if it is linearly independent and generates $E$, i.e, $\text{span}((x-2),(x-2)^2,(x-2)^3)=E$. Proving that $\mathcal{B}$ is linearly independent shouldn't be hard. Let $\alpha ,\beta ,\gamma\in \mathbb{R}$ such as $\alpha (x-2)+\beta (x-2)^2+\gamma (x-2)^3=0$. Expand this and write it in the form $Ax^3+Bx^2+Cx+D=0$ and so $A=B=C=D=0$. Then you must find that $\alpha=\beta=\gamma=0$. Another easier way is to differentiate $\alpha (x-2)+\beta (x-2)^2+\gamma (x-2)^3=0$ as many ways necessary to get rid of $\alpha$ and $\beta$ but still have $\gamma$ and so you'll find that $\gamma =0$ and so on...
Proving that $\mathcal{B}$ spans $E$ isn't really hard. Let $P\in E$ . Then $P(x)=(x-2)(ax^2+bx+c)$ for some $a,b,c\in\mathbb{R}$ (you should use $P(2)=0$ and $\deg(P)\le3$ to see why). Let's focus on Q(x)=$ax^2+bx+c$ and see if it can be written as a linear combinaison of elements of $\mathcal{B}$. First put $y=x-2$. Then $Q(x)=a(y+2)^2+b(y+2)+c$. If you expand it, it'll be of the form $Q(x)=my^2+ny+q$ for some $m,n,q\in \mathbb{R}$ and $y=x-2$. Back to $P(x)=(x-2)Q(x)$ conclude that $P$ is a linear combinaison of vectors of $\mathcal{B}$.

This is an important result:

Let $\mathcal{F}=\{P_1,\,P_2,\,\dots ,P_n\}$ be a family of nonzero vectors of $\mathbb{R}[x]$. Prove that if they all have different degrees then $\mathcal{F}$ is linearly independent.

Try to prove it! You can do as I proposed to prove that $\mathcal{B}$ is linearly independent.
