Find the dimension and a basis of $V$ - the set of all polynomials over $\mathbb{R}$ of degree at most $3$ that vanish at the point $x = 1$.
So, I know that the dimension would be $4$ since the dimension of a set of polynomials at degree most $k$ would be $k+1$.(?). But the vanishing on $1$ part confuses me. Like how am I supposed to act with this fact? Also how do I find the basis?
Let $W = P_3(\Bbb{R})$ be the vector space of real-valued polynomials. We consider this a subspace of the polynomials as a subspace of the vector space of all functions $\Bbb{R} \to \Bbb{R}$, which is to say, we equip $W$ with the operations: \begin{align*} (p \color{red}+ q)(x) &:= p(x) + q(x) \\ (\lambda \color{red}\cdot p)(x) &:= \lambda \cdot p(x), \end{align*} where $p, q \in W$, $\lambda \in \Bbb{R}$, and $x$ is a dummy variable ranging over all $\Bbb{R}$. The two operations highlighted in red are the operations being defined here. The red $\color{red}+$ is a sum of two functions, where as the black $+$ is the sum of two numbers in $\Bbb{R}$. Similarly, the red $\color{red}\cdot$ is scalar multiplication of a real number (a scalar) and a function (a vector), whereas the black $\cdot$ is the standard multiplication on $\Bbb{R}$.
Recall the definition of a polynomial and its degree. Essentially, to say that $p$ is a polynomial of degree $3$ or less is to say that there exist some scalars $a_0, a_1, a_2, a_3 \in \Bbb{R}$ such that $$p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3.$$ That is, $B = (1, x, x^2, x^3)$ spans $W$, basically by definition of $W$. If we can show that these functions are linearly independent, then they are a basis for $W$, and $W$ has dimension $4$.
There are a few ways to do this, most of which begin with assuming you have $a_0, a_1, a_2, a_3 \in \Bbb{R}$ such that $$a_0 \cdot 1 + a_1 \cdot x + a_2 \cdot x^2 + a_3 \cdot x^3 \equiv 0, \tag{1}$$ by which I mean the two expressions are equal for all $x$. Naively, we can simply substitute in some specific values for $x$ into $(1)$, say $x = -1, 0, 1, 2$, to obtain a system of linear equations \begin{align*} a_0 - a_1 + a_2 - a_3 &= 0 \\ a_0 &= 0 \\ a_0 + a_1 + a_2 + a_3 &= 0 \\ a_0 + 2a_1 + 4a_2 + 8a_3 &= 0. \end{align*} Solving these four linear equations yields a unique solution of $a_0 = a_1 = a_2 = a_3 = 0$, proving $B$ is linearly independent.
There is a more elegant way too. Since $(1)$ is an equality of (infinitely differentiable) functions, we can differentiate both sides. In fact, differentiating both sides $3$ times yields $$6a_3 = 0 \implies a_3 = 0.$$ Differentiating $(1)$ twice then yields $$6a_3 x + 2a_2 = 0 \implies a_2 = 0.$$ Differentiating $(1)$ once implies $$3a_3x^2 + 2a_2 x + a_1 = 0 \implies a_1 = 0,$$ and finally, $(1)$ itself now implies $a_0 = 0$. The beauty of this method is that it extends to arbitrary polynomials, not just ones of degree $3$ or less. Either which way, $B$ is a basis for $W$.
(Some people will prove the linear independence of $B$ by equating coefficients, but I consider this a circular proof. Equating coefficients is another way of saying the linear combinations of $B$ are unique, a property equivalent to the linear independence of $B$. In order to show that equating coefficients is valid, you need a technique like the above.)
Now, on to my hint. I suggested looking at the basis $C = (1, x - 1, (x - 1)^2, (x - 1)^3)$. There are a number of ways to show that $C$ is a basis for $W$, but I'm going to do it the naive way.
Consider an arbitrary polynomial $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ in $W$. Let $u = x - 1$. Then $x = u + 1$, and \begin{align*} p(x) &= p(u + 1) \\ &= a_0 + a_1 (u + 1) + a_2 (u + 1)^2 + a_3 (u + 1)^3 \\ &= (a_0 + a_1 + a_2 + a_3) + (a_1 + 2a_2 + 3a_3)u + (a_2 + 3a_3)u^2 + u^3 \\ &= (a_0 + a_1 + a_2 + a_3) + (a_1 + 2a_2 + 3a_3)(x - 1) + (a_2 + 3a_3)(x - 1)^2 + (x - 1)^3. \end{align*} Hence, $p \in \operatorname{span}(C)$, so $C$ spans $W$.
On the other hand, suppose $$a_0 \cdot 1 + a_1 \cdot (x - 1) + a_2 \cdot (x - 1)^2 + a_3 \cdot (x - 1)^3 \equiv 0.$$ Then, expanding $$(a_0 - a_1 + a_2 - a_3) \cdot 1 + (a_1 - 2a_2 + 3a_3) \cdot x + (a_2 - 3a_3) \cdot x^2 + a_3 \cdot x^3 \equiv 0.$$ By the linear independence of $B$, we get \begin{align*} a_0 - a_1 + a_2 - a_3 &= 0\\ a_1 - 2a_2 + 3a_3 &= 0 \\ a_2 - 3a_3 &= 0 \\ a_3 &= 0, \end{align*} which clearly implies $a_0 = a_1 = a_2 = a_3 = 0$, i.e. $C$ is linearly independent. Hence, $C$ is also a basis for $W$.
(Note that we could have taken a shortcut. Since $B$ is a basis and has $4$ vectors, $W$ is $4$-dimensional, so we could have showed either $C$ is linearly independent or $C$ is spanning, and obtained the other for free!)
Finally, we can use this basis $C$ to find a basis for $V$. Note that each vector in $C$ belongs to $W$, except $1$ (since $1$ is a function that constantly takes the value $1$, hence it doesn't vanish at $x = 1$, or indeed any other point). If we remove $1$, we get a (necessarily linearly independent) list of vectors $$D = (x - 1, (x - 1)^2, (x - 1)^3).$$ All three vectors in $D$ belong to $V$, so $V$ has dimension at least $3$. It is a strict subspace of $W$ (e.g. the constant function $1$ is in $W$, but not $V$), so the dimension is strictly less than $4$. Thus, $$\operatorname{dim} V = 3.$$ Hence, any linearly independent set of $3$ vectors from $V$ (e.g. $D$) will be a basis. Thus, $D$ is indeed a basis for $V$.