Using Rank-Nullity Theorem, find dimension and basis

Using Rank-Nullity Theorem, find dimension and basis for the Subspace $W$ of $\mathbb{Q}_{4} [x]$ consisting of all the polynomials $a_{0}+a_1x+a_2x^2+a_3x^3+a_4x^4$ such that $a_1+a_2+a_3+a_4=0$.

Not sure how to use the theorem with this problem and very unsure what steps need to be taken.

Thank you in advance for the help.

The Rank-Nullity Theorem doesn't do anything to help construct a basis. But you could use it to determine the dimension of $W$. Let $$T: \mathbb{Q}_4[x] \to \mathbb{Q}$$ be the transformation defined by $$T(a_0 + a_1x + a_2x^2 + a_3 x^3 + a_4 x^4) = a_1 + a_2 + a_3 + a_4.$$

Then $W = ker(T)$ and rank($T) = 1$.

The Rank-Nullity Theorem says that dim$(W)= \text{nullity}(T) = \text{dim}(\mathbb{Q}_4[x]) - \text{rank}(T) = 5-1 = 4$.

Now we need to construct a basis of $W$. That is we need $4$ vectors in $W$ that span.

If $f(x) = a_0 + a_1x + a_2 x^2 + a_3x^3 + a_4x^4 \in W$, then $a_1 = -a_2 - a_3 - a_4$. Therefore

\begin{align*} f(x) &= a_ 0 + (-a_2 - a_3 - a_4)x + a_2 x^2 + a_3x^3 + a_4x^4 \\ &= a_ 0(1) + a_2(x^2 - x) + a_3(x^3 - x) + a_4(x^4-x) \in \text{span}(1, x^2-x,x^3-x,x^4-x) \end{align*}

$\{1, x^2-x,x^3-x,x^4-x\}\subseteq W$ is a spanning set of $W$ of the correct cardinality and is thus a basis.

Hint:

Consider the linear map \begin{aligned}[t]\varphi\colon\mathbf Q_4[x]&\longrightarrow \mathbf Q,\\a_0+\dots +a_4x^4&\longmapsto a_1+\dots+a_4.\end{aligned}

This map is surjective and $\;W=\ker \varphi$.