Find a basis for the space of cubic polynomials $p$ such that $p(3) = 0$ What will be the basis for the space of cubic polynomials $p$ such that $p(3) = 0$?
I know that: A natural basis for the vector space of cubic polynomials  is $p(3)$ is $\langle 1, x, x^2, x^3 \rangle$. 
 A: One way to arrive at the basis proposed by André is the following: if $p(x)= ax^3 + bx^2 +cx +d$ has $3$ as a root, then
$$
0 = p(3) = 27a + 9b + 3c + d \ \qquad \Longrightarrow \qquad \ d = -27a - 9b -3c \ .
$$
This means that for polynomials of degree less than or equal to $3$ having $3$ as a root, you can choose, for instance, coefficients $a,b,c$ at will, but then you have no other choice for $d$ than $-27a-9b-3c$. So these polynomials look like:
$$
p(x)= ax^3 + bx^2 +cx -27a-9b-3c = a(x^3-27) + b(x^2 - 9) - c(x-3) \ .
$$
Remember that you can choose any $a,b,c$? Hence, this last expression means that all polynomials having degree less than or equal to $3$ and $3$ as a root are linear combinations of André's basis:
$$
x^3-27,\  x^2 - 9,\ x-3 \ .
$$
So, these ones generate all your polynomials and now you just need to prove that they're linearly independent.
A: We want to find a basis for the vector space $V$ of all polynomials $p(x)$ of degree less than or equal to $3$ such that $p(3)=0$.  (I will assume that the vector addition is ordinary addition of polynomials, and that the multiplication by scalars is the usual one. Maybe you are expected to prove that $V$ really is a vector space. I leave that part, if it is needed, to you.)
It is easy to see that $V$ is a proper subspace of the space of all polynomials $p(x)$ of degree less than or equal to $3$ (after all, $1$ is not in $V$). In particular, the dimension of $V$ is less than $4$.
We will show that 
$$\{x-3, x^2-9, x^3-27\}$$ 
is a basis of $V$.
It is easy to verify that the vectors in our proposed basis are linearly independent (that part is left to you).  They all vanish at $x=3$. It follows that any linear combination $a(x-3)+b(x^2-9)+c(x^3-27)$ of them must also vanish at $3$. 
There are $3$ vectors in our proposed basis.  Thus they generate a subspace of $V$ of dimension $3$. As we observed earlier, $V$ has dimension less than $4$. This shows that our $3$ vectors generate all of $V$.
There are many other choices of basis: any $3$ linearly independent polynomials of degree $\le 3$ that vanish at $3$ will do the job.
A: Take a basis of the space of degree at most $2$ polynomials, and multiply it by $x-3$. 
More precisely, a tuple $(p_1,p_2,\dots)$ of polynomials is a basis that fits the bill, if and only if each $p_i$ is divisible by $x-3$, and the tuple of quotients is a basis of the space of degree at most $2$ polynomials. 
A: The subset of $\mathbb{R}[X]$, $\mathbb{R}_3[X]=\{P\in \mathbb{R}[X], deg(P)\leq 3\}$ is a vectorial subspace of dimension 4 a basis of which is $1$, $x$, $x^2$, $x^3$. $\phi:\mathbb{R}_3[X]\rightarrow \mathbb{R}_3[X], \phi(P)= P(3)$ is a linear form on this subspace. The set we are considering is its kernel and therefore is a subspace of dimension $4-1=3$. The set $\{x-3,(x-3)^2,(x-3)^3\}$ is a free family of three polynomials such that $\phi(P)=0$ and thus the basis we are looking for.
