Find the possible signatures of the bilinear forms

Question

Find the possible signatures of the following bilinear forms:

1. The bilinear form $\phi:\mathbb R^n\times\mathbb R^n\to\mathbb R$ given by $\phi(x,y)=x^Tp(A)y$ where $p(t)=t^2+bt+c$ is a quadratic with no real root and $A$ is a symmetric $n\times n$ matrix.

2. The bilinear form $\phi:V\times V\to \mathbb R$ is defined on a real vector space $V$ with $\dim(V)=5$ and for every $v\in V$ there exists $v_1\in V$ such that $\phi(v,v_i)\neq0$

My attempts are as follows. Please remark whether they are correct or not.

1. The given condition on $p$ implies that $p(A)$ is a symmetric non-singular matrix. Hence it can be converted to a diagonal matrix of the form which contains no $0$ in its diagonal. In particular, as $p(t)$ does not have any real root, it must happen that $p(t)>0$ for all $t\in\mathbb R$, which further implies that $p(A)$ is positive definite. Hence the number of positive diagonal entries in the diagonal form must be equal to $n$. So the only possible signature is $n-0=n$.
2. Define $t_v:V\to\mathbb R$ by $t_v(w)=\phi(t,w)$ for a fixed $v\in V$ and any $w\in V$. Then $t$ is non-null as I can always get a $w$ such that $t_v(w)\neq0$. Now, observing that $t$ is linear, we invoke the well known theorem that $\dim(V)=Rank(t)+Nullity(t)$. Also, obviously, $Rank(t)=0$ or $1$ as $\mathbb R$ is a scalar field. $Rank(t)=0$ implies the range of $t$ is zero, which is false by hypothesis, so $Rank(t)=1$. Hence $Nullity(t)=4$. Hence we can find $4$ linearly independent vectors in $V$, viz. $u_1,u_2,u_3,u_4$ such that $t_v(u_i)=0$ for $1=1,2,3,4$ i.e. $\phi(v,u_i)=0$ for $i=1,2,3,4$. Now $v$ was arbitrary and this holds for any $v\in V$. Our matrix for the bilinear form $\phi$ has been constructed using a basis, and now, after obtaining $u_1,u_2,u_3,u_4$ as basis vectors for the Null Space of the matrix of the bilinear form, and noting that the choice of $u_1,u_2,u_3,u_4$ was independent of $v$, we can say that the matrix of the bilinear form is diagonalizable with $4$ diagonal entries $0$ and remaining one entry nonzero. Hence signature = $-1$ or $1$.