= Solution
Let $f_1,f_2\in V^*$ satisfy $f_1(v)f_2(v)=0$ for every <vector> $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne0$. For every $v\in\ker f_1$ and every <real number> $t$,
$$
f_1(u+tv)=f_1(u)\ne0,
$$
so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Since
$$
V=\mathbb Ru+\ker f_1,
$$
we obtain $f_2=0$. Thus one of the two <linear functionals> is zero.
<Sylvester's law of inertia> says that, in a suitable <basis>, every real <quadratic form> is
$$
q=x_1^2+\cdots+x_p^2-x_{p+1}^2-\cdots-x_{p+m}^2.
$$
The <rank> is $r=p+m$, and the <signature> is $\sigma=p-m$; both are independent of the chosen basis.
If $q=f_1f_2$, its <polarization identity>[polar form] has image contained in $\operatorname{span}\{f_1,f_2\}$, so $r\leq2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma=0$. If $r=1$, then $|\sigma|=1$. In every case,
$$
r+|\sigma|\leq2.
$$
Conversely, this inequality leaves only
$$
q=0,\qquad q=\pm x_1^2,qquad q=x_1^2-x_2^2,
$$
up to a <change of basis>[linear change of coordinates]. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore
$$
\boxed{q=f_1f_2\quad\Longleftrightarrow\quad r+|\sigma|\leq2}.
$$
Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectors
$$
e_1,\ldots,e_p,\quad f_1,\ldots,f_m,\quad z_1,\ldots,z_k
$$
with $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the <radical of a bilinear form>[radical]. The $p+m+k$ vectors
$$
e_1+f_1,\ e_1-f_1,\ e_i+f_1\ (i\geq2),\ e_1+f_j\ (j\geq2),\ z_l
$$
are linearly independent and all satisfy $q(v)=0$. They form the required basis of <isotropic vector>[isotropic vectors].
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