Vector space of polynomials with given root Let $\mathbb{R}[t]_{\leq n}$ represent the vector space of polynomials (with coefficients in $\mathbb{R}$) whose degree is at most $n$. 
$\forall a \in \mathbb{R}$, let $U_a = \{P(t) \in \mathbb{R}[t] \mid P(a) = 0 \}$
1) Find a basis of $S = U_a \cap \mathbb{R}[t]_{\leq n}$ for all $a \in \mathbb{R}$
For all $P(t) \in S$, $ P(t) = \lambda_0 + \lambda_1t + \lambda_2 t² + \cdots + \lambda_nt^n$ for $\lambda_i \in \mathbb{R}, \forall i$. We also know that $P(a) = 0$ so we can write:
$\sum_{i=0}^{n}(\lambda_ia^i) = 0$
But beyond this, I'm not sure how to proceed..
2) Show that $(U_3+U_4) \cap \mathbb{R}[t]_{\le n} = \mathbb{R}[t]_{\le n}$.
I'm stumped. Wouldn't this mean that $\forall p \in \mathbb{R}[t]_{\leq n}$, p can be written  $p=Q(t)(t-3)+Q'(t)(t-4)$ with $deg(Q)$, $deg(Q') < deg(P)$  ?
 A: 1) Multiply the standard basis $1,t,t^2,..$ by $t-a$.
2) Well, I see the point now. If it wants to
 mean, with $deg(Q),deg(Q')\le n$, then it easily follows from $(t-3)-(t-4) = 1$, as $P(t)=P(t)\cdot\big( (t-3)-(t-4) \big)$.
But, in this formulation (even if you rather meant $deg(Q),deg(Q')<n$) it is not true for the constants, $n=0$. 
For higher $n$, it seems it is still true, as $t=4(t-3)-3(t-4)$, then for the term containing $t^k$ in $p$ you can use $t^{k-1}\cdot \big(4(t-3)-3(t-4)\big)$, and $(t-3)-(t-4)$ if $k=0$.
A: Hints: you're going to need the very nice and sometimes pretty helpful
Proposition: if $\,V_\Bbb F\,$ is a vector space and the $\,f:V_\Bbb F\to\Bbb F\,$ is a non-zero linear functional , then $\,\ker f\,$ is a maximal subspace of $\,V_\Bbb F\,$ , meaning: 
$$\forall\,x\in V_\Bbb F\,\,,\,x\notin\ker f\Longrightarrow V_\Bbb F=Span\,\{\ker f, x\}\,$$
1) Show that 
$$U_a=\ker f_a\,\,,\,\text{with}\,f_a:\Bbb R[t]_{<n}\to\Bbb R\,\,\,\text{defined by}\,\,f_a(p(t)):=p(a)$$
2) Just show that $\,U_3+U_4=\Bbb R[t]_{<n}\,$ ...yes, this would mean what you think but without the reduntant condition on the degrees.
