Hint - Find a basis for $U=$ {$p \in P_4(R) : p(6)=0$} Let $P_4(R)$ denote the set of all polynomials of at-most degree $4$. I was attempting to find a basis of $U=$ {$p \in P_4(R) : p(6)=0$}. I do not want a solution, but just a hint as to where to look. The following was my approach.
Because $p(6)$ must equal $0$. We know that the polynomial will have a factorisation such as $(x-6)(x-a)(x-b)(x-c)$. Multiplying throughout this polynomial, we see that at the end there is always a multiple of $6$. Hence, the basis would consist of the number 6. But because this polynomial can have degree: 1, 2, 3, 4. We would also have to add the vectors
$x, x^2, x^3, x^4$(Abuse of notation here, what I mean is that they are all function. $f_j(x)=x^j$).
So, the basis would consists of $6, x, x^2, x^3, x^4$. We might be able to change this basis up a bit. We want that $a_0(6)+a_1(6)+a_2(6)^2+a_3(6)^3+a_4(6)^4=0$.
At this point I am stuck. Because in the next question, it asks to extend this basis to a basis of $P_4(R)$. But this basis would also be a basis of $P_4(R)$, as any linearly independent list of the right length is a basis of that vector space.
Could someone give any hints as to what I should be looking for?
 A: HINT
If $p\in U\leq\textbf{P}_{4}(\mathbb{R})$, it can be written as $p(x) = a + bx + cx^{2} + dx^{3} + ex^{4}$. Hence we conclude that:
\begin{align*}
p(6) = 0 & \Rightarrow a + 6b + 36c + 196d + 1296e = 0\\\\
& \Rightarrow a = -(6b + 36c + 196d + 1296e)
\end{align*}
Now you can express the elements of $U$ in terms of four parameters.
Can you proceed from here?
A: Define $T: P_4(\Bbb{R})\to \Bbb{R}$ by
$$T_6(p) =p(6) $$ $\forall p\in P_4(\Bbb{R})$
Then

*

*$T_6$ is a linear map .


*$T_6$ is surjective.


*$\ker T_6=\{p  \in  P_4(\Bbb{R}): p(6)=0\}$
Then by Rank -Nullity theorem,
$\dim(\ker(T_6)) =\dim(P_4(\Bbb{R})) -\dim(R(T_6)) =5-1=4$
A: If you multiply out,  you get $x^4-(6+a+b+c)x^3+(6a+6b+6c+ab+ac+bc)x^2-(6ab+6bc+6ac+abc)x+6abc$.
In the standard basis,  $\{x^4,x^3,x^2,x,1\}$, that's the vector $(1,-(6+a+b+c),6a+6b+6c+ab+ac+bc,-(6ab+6bc+6ac+abc),6abc)$.
Now choose $a,b,c=0$ and get $(1,-6,0,0,0)$.  Next choose $a=1,b=c=0$ to get $(1,-7,6,0,0)$.  Let $a=1,b=1,c=0$, get $(1,-7,13,-6,0)$.  And $a=b=c=1$ for $(1,-9,21,-19,6)$.
This gives one basis.   Notice there are $4$ basis vectors,  so the dimension is $4$.

Now to extend,  notice that there's no way,  with this basis,  to get a nonzero constant polynomial (or, any polynomial that isn't zero at $6$, for that matter).
