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Question :
Find a basis for $V=\{p(x) \in \mathbb R_n[x]:p(1)=p(2)=0\}$ (Also find the dimension of $V$ ).

Note : By $\mathbb R_n[x]$, I mean all of the polynomials of at most degree $n$ with real coefficients. (Like $f(x)=c_0+c_1x+c_2x^2+\dots+c_nx^n$)

My problem on solving this :
I've solved many similar questions when $n$ was small (like $2$ or $3$). But in this case, the ways i used (like elementary row operations) are not useful. I can't for example find two things like $x_i$ and $x_j$ and write all the other things with respect to them.
What should i do with these equations?!
$p(1)=c_0+c_1+\dots+c_n$
$p(2)=c_0+2c_1+4c_2+\dots+2^nc_n$

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    If $p(1)=p(2)=0$, in particular it means either it is $0$ polynomial, or $p(x)$ is a multiple of $(x-1)(x-2)$ – 2017-02-10
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    The dimension is easy to find for $n\geq 1$. Take the map $\mathbb{R}_n[x]\to \mathbb{R}^2$ sending each $p(x)$ to $\big(p(1),p(2)\big)$. This map is surjective with kernel $V$. Therefore, $$\dim_\mathbb{R}(V)=\dim_\mathbb{R}\big(\mathbb{R}_n[x]\big)-\dim_\mathbb{R}\big(\mathbb{R}^2\big)=(n+1)-2=n-1\,.$$ For a basis, use @user160738's hint. – 2017-02-10
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    @Batominovski i still don't get it... how should i find the basis? (But i understand both of your comments) – 2017-02-10
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    If $p(x)\in V$, then what are restrictions on the possible values of $\frac{p(x)}{(x-1)(x-2)}$? – 2017-02-10
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    Following the different indications that have been given, a basis is the set of $n-1$ polynomials $x^k(x-1)(x-2)$ for k=0,1,2...(n-2)$. – 2017-02-10

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