Finding dimension of subspace 
I know that any polynomial in subspace $W$ must have $(x-1)$ as factor so that $p(1)=0$
But I don't understand how $p'(2)=0$ can be incorporated.
Thankful for any kind of help.
 A: A general element $p(x)\in V$ has the form 
$$p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$$
for some $a,b,c,d,e,f\in\mathbb{R}$. 
The condition $p(1)=0$ means
$$a+b+c+d+e+f=0.$$
The condition $p'(2)=0$ means
$$80a+32b+12c+4d+e=0.$$
So you get that the subspace $W$ consists of the polynomials for which the coefficients are the solutions to the following set of two linear equations in six variables:
$$\begin{cases} 
a+b+c+d+e+f=0\\
80a+32b+12c+4d+e=0
\end{cases}$$
Can you take it from here?
A: The easier way is using coordinates so the conditions $p(1)=0$ and $p'(2)=0$ transform into two linearly independent equations (as Brian Fitzpatrick shows). Then apply the next formula dim$(W)$=dim$(V)$-$k$, where $k$ is precisely the number of linearly independent equations used to define $W$. So dim$(W)=6-2=4$.
The idea is that each linear independent equation used to define a subspace, subtracts one dimension from it.
A: You can define a linear transformation
$$ T:\mathbb{R}_5[x]\longrightarrow \mathbb{R}^2$$
such that $p \longmapsto (p(1),p'(2))$. It is to see that $T$ is surjective. In particular $kerT=W$; then by the dimension theorem we have:
$$dim(V)=dim( ker T) + dim( Im T)$$
so $6=dim(kerT)+2 \quad \Longrightarrow dim(ker T)=4$.
