= Solution
The <dual space> $V^*$ is the <vector space> of <linear maps> $V\to\mathbb F$, with pointwise operations. For the given <basis> define coordinate forms $e^i$ by $e^i(\sum_ja_je_j)=a_i$. These are linear and satisfy $e^i(e_j)=\delta_{ij}$. If $\sum_ic_ie^i=0$, evaluation at $e_j$ gives $c_j=0$. For every <linear form> $\phi$,
$$
\phi(v)=\sum_i\phi(e_i)e^i(v),
$$
so the coordinate forms span the dual and form its <dual basis>.
In the <polynomial> example write $p(X)=a+bX+cX^2$. Then
$$
\phi_1(p)=a,\qquad\phi_2(p)=a+b+c,\qquad\phi_3(p)=a-b+c.
$$
These recover all three <coefficients>, so they span $V^*$ and, since the dual has <dimension> three, form a <basis>. Explicitly the <basis> dual to $1,X,X^2$ is
$$
\boxed{e^1=\phi_1,\qquad e^2=\frac{\phi_2-\phi_3}2,\qquad e^3=\frac{\phi_2+\phi_3}2-\phi_1.}
$$
Evaluation on the three monomials verifies the dual-<basis> identities directly.
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