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I am looking to solve the following problem regarding polynomials:

Suppose that all of the roots $(r_1, r_2, r_3) \in \mathbb{R}^3$ of a linear form $\sum_{k=1}^3 g_k x_k$ are roots of a quadratic form $\sum_{i,j=1}^3 b_{ij}x_ix_j$. How do you show that exists another linear form $\sum_{l=1}^3 h_l x_l$ such that $$ \sum_{i,j=1}^3 b_{ij}x_ix_j = \left( \sum_{k=1}^3 g_k x_k \right) \left( \sum_{l=1}^3 h_l x_l \right). $$

Could someone give me some advice/help?

I know that if the forms were in two variables, it would be "trivial". From this post Do polynomials in two variables always factor in linear terms?, when there are more than two terms, it gets harder.

Shurik G.
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    I don't see why $x_1, x_2,$ and $x_3$ are roots of $g_1x_1 + g_2x_2 + g_3 x_3$; indeed, I don't even know whether the $x_i$ are meant to be constants or indeterminates. Can you clarify, perhaps giving names to some functions or something? – John Hughes Jul 02 '19 at 13:43
  • @JohnHughes we suppose that they are roots of a multivariate polynomial. What do you not understand? – Shurik G. Jul 02 '19 at 13:46

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Change coordinates so that the vector $g = (g_1, g_2, g_3)$ is $(1, 0, 0)$.

Now we have a quadratic $B(x) = \sum b_{ij} x_i x_j$ that's zero on the whole $x_1$-axis. The first Lemma in Chapter 3 of Samuel's Projective Geometry says (with some parts omitted)

Lemma: let $P(x_1, \ldots, x_n)$ be a homogeneous quadratic polynomial over a field $K$. Then

(a) [...]

(c) if $P(0, a_2, \ldots, a_n)$ for every $a = (a_2, \ldots, a_n) \in K^{n-1}$, then $P$ is a multiple of $x_1$.

Applying that to our situation, we see that $B(x) = x_1 H(x)$, where $H$ is necessarily homogeneous of degree $1$, i.e., a linear form, and we're done.

The proof Samuel gives is this:

Write $P(x_1, \ldots, x_n) = c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + P_2(x_2, \ldots, x_n)$, where $P_i$ is homogeneous of degree $i$. Now evaluating this at $(0, a_2, \ldots, a_n)$ gives $0 = 0 + 0 P_1(a_2, \ldots, a_n) + P_2(a_2, \ldots, a_n)$, hence the homogeneous quadratic polynomial $P_2$ is zero. But then \begin{align} P(x_1, \ldots, x_n) &= c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + P_2(x_2, \ldots, x_n)\\ &= c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + 0\\ &= x_1 \left( c_1 x_1 + P_1(x_2, \ldots, x_n)\right). \end{align}

[I've put words in Samuel's mouth here; he basically says "write $P$ this way; then it's clear that $P_2$ is zero."]

John Hughes
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