I was asked to find the signature of the quadtratic form $Q(p)=p(1)p(2)+p(3)p(4)$ where $p$ is a polynomial in $\mathbb R_n[x]$
I tried doing it via finding the symmetric matrix that $Q$ corresponds to, and I did that by trying to find out what is the image of a general polynomial, $$p(x)=\sum_{i=0}^n \alpha_ix^i$$
$$Q(p)=\sum_{i=0}^n \alpha_i \sum_{j=0}^n \alpha_i2^i + \sum_{k=0}^n \alpha_i3^i\sum_{r=0}^n \alpha_i4^i$$
How do we go on from here?
We could also try finding the maximal subspace for which $Q$ is definite positive / definite negative, but that doesn't seem easier.
When $\mathbf{n=0}$, the quadratic form has signature $\mathbf{(1,0)}$ since $Q(1)=2>0$.
When $\mathbf{n=1}$, the quadratic form has matrix $$\mathrm{Mat}(\beta,(2-X,X-1))= \begin{pmatrix} 0&\frac12(1+0+6+3)\\ \frac12(0+1+6+3)&0 \end{pmatrix}=\begin{pmatrix} 0&5\\ 5&0 \end{pmatrix}$$ and thus signature $\mathbf{(1,1)}$.
When $\mathbf{n=2}$, the matrix of $Q$ w.r.t. the basis $\left(\frac{(X-2)(X-3)}{(1-2)(1-3)},\frac{(X-1)(X-3)}{(2-1)(2-3)},\frac{(X-1)(X-2)}{(3-1)(3-2)}\right)$ is $$\begin{pmatrix} 0&\frac12&\frac12\\ \frac12&0&-\frac32\\ \frac12&-\frac32&1 \end{pmatrix}$$ Since the subspace $\Bbb R$ is positive for $Q$, and the determinant of the above matrix equals $-1$, $Q$ necessarily has signature $\mathbf{(2,1)}$.
When $\mathbf{n\geq 3}$, consider the basis $(\phi_1,\phi_2,\phi_3,\phi_4,\dots,\phi_n,\phi_{n+1})$ of $(\Bbb R_n[X])^*$ where $\phi_i:\Bbb R_n[X]\to\Bbb R$ is the linear form that maps $P$ to $P(i)$. By definition, $Q(P)=\phi_1(P)\phi_2(P)+\phi_3(P)\phi_4(P)$. Its antedual basis is the basis $(L_i)_{i=1,\dots,n+1}$ of $\Bbb R_n[X]$ where $$L_i=\prod_{j\neq i}\,\frac{X-j}{i-j}$$ Let $\beta$ be the polar form of $Q$, that is, for all $A,B\in\Bbb R_n[X]$, $$\beta(A,B)=\frac12\Big(\phi_1(A)\phi_2(B)+\phi_2(A)\phi_1(B)+\phi_3(A)\phi_4(B)+\phi_4(A)\phi_3(B)\Big)$$ the matrix of $\beta$ w.r.t. this basis is $$\mathrm{Mat}(\beta,(L_i))= \begin{pmatrix} 0&\frac12\\ \frac12&0\\ &&0&\frac12\\ &&\frac12&0\\ &&&&0\\ &&&&&\ddots\\ &&&&&&0 \end{pmatrix}$$ Thus the signature of $Q$ is the same as that of the quadratic form defined by the symmetric matrix $$\begin{pmatrix} 0&1\\ 1&0\\ &&0&1\\ &&1&0\\ \end{pmatrix}$$ that is, $\mathbf{(2,2)}$. To prove this, just notice that the matrix of $\beta$ in the basis $\mathcal B=(L_1+L_2,L_1-L_2,L_3+L_4,L_3-L_4,L_5,\dots,L_{n+1})$ is $$\mathrm{Mat}(\beta,\mathcal B)= \begin{pmatrix} 1\\ &-1\\ &&1\\ &&&-1\\ &&&&0\\ &&&&&\ddots\\ &&&&&&0 \end{pmatrix}$$