Trouble with problem involving finding a basis for the set Having a lot of trouble with these problems.Can someone show me how to show the set of vectors I have found spans $U$?
Problem Let $U=\{p \in \mathcal{P}_4(F):p(6)=0\}$. 
(a).Find a basis of $U$.
Attempt:
(a). Using methods from this forum to help me I found a basis to be 
$(x-6),(x^2-36),(x^3-216),(x^4-1296)$
The vectors are linearly independent since
$a(x-6)+b(x^2-36)+c(x^3-216)+d(x^4-1296)=0$
Thus $6a-36b-216c-1296d+ax+bx^2+cx^3+dx^4=0$
$\implies a=b=c=d=0$
Now I cannot figure out how to show that the set spans $U$ any help?
How am I supposed to know what a generic vector in $U$ is so I can write the generic vector as a linear combination of my basis?
 A: Let $p(x)=ax^4+bx^3+cx^2+dx+e \in U$. Then $p(6)=0 \implies 1296a+216b+36c+6d+e=0$. Thus 
$e=-(1296a+216b+36c+6d)$. So
\begin{align*}
p(x) & =ax^4+bx^3+cx^2+dx-(1296a+216b+36c+6d)\\
&=a(x^4-1296)+b(x^3-216)+c(x^2-36)+d(x-6).
\end{align*}
Thus any $p(x) \in U$ is a linear combination of the vectors in your basis.
A: Depending on the method that you used to generate this basis, you should already know that these vectors span it. However, here are a couple of ways to verify this.  
$U$ is the set of polynomials annihilated by the “evaluate at $6$” functional. By a variant of the Rank-Nullity theorem, this implies that $U$ is four-dimensional. All four polynomials in your basis clearly vanish at $x=6$ and they’re obviously linearly independent, so they must be a basis of $U$.  
Alternatively, $U$ is the set of polynomials of degree at most four that have $6$ as a root. Every such polynomial is of the form $(x-6)(ax^3+bx^3+cx+d)$. Factoring $x-6$ out of your proposed basis vectors, the problem reduces to showing that $\{1,x+6,x^2+\cdots,x^3+\cdots\}$ is a basis of $\mathcal P_3(F)$, but these four polynomials are obviously linearly independent, so you’re done. Notice that I didn’t bother to work out all of the quotients of division by $x-6$. It’s enough to know that the four polynomials are all of different degree to determine that they’re independent.
