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Let $n \geq 10$ and set $\mathbf{y} = (y_1,\ldots,y_n)$. Let $Q_1(\mathbf{y}),\ldots,Q_5(\mathbf{y})$ be non-zero quadratic forms with integer coefficients such that the cubic form $x_1Q_1(\mathbf{y})+\ldots+x_5Q_5(\mathbf{y})$ is non-degenerate in the $x_i$ and irreducible over $\overline{\mathbf{Q}}$. Then I would like to show that the set $$ V = \left\{(x_1,\ldots,x_5) \in \mathbf{A}_{\mathbf{Q}}^5 : x_1Q_1(\mathbf{y})+\ldots+x_5Q_5(\mathbf{y}) \text{ is reducible over }\mathbf{Q}\right\} $$ is contained in a proper subvariety of $\mathbf{A}_{\mathbf{Q}}^5$.

This is true, for example, if the generic fibre of the morphism $(\mathbf{x},\mathbf{y}) \to \mathbf{x}$ from $X$ to $\mathbf{A}_{\mathbf{Q}}^5$ is geometrically irreducible, where $X$ is the cubic hypersurface cut out by the equation $\sum_{i=1}^5x_iQ_i(\mathbf{y}) = 0$ (see https://stacks.math.columbia.edu/tag/0559). However, I am not sure how to proceed if this is not the case.

Any help to show that $V$ is contained in a proper subvariety would be much appreciated.

Note: An earlier version of this question did not specify that the $Q_i$ are non-zero. If this is the case, $V$ is Zariski-dense. Thanks to Jason Starr for pointing this out.

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    $\begingroup$ That is not true. Let $Q_1 = y_1^2$, let $Q_2=-y_2^2$, and let $Q_3=\dots=Q_5 = 0$. Then $V$ is the subset of $\mathbf{A}^5_{\mathbf{Q}}$ such that $x_1$ equals $0$ or $x_2/x_1$ is a square. That is a Zariski dense subset. $\endgroup$
    – Jason Starr
    Jan 30 at 11:22
  • $\begingroup$ @JasonStarr thanks for your answer. Actually, in my application, none of the quadratic forms Q_i can vanish (I have edited my question accordingly). Apologies. $\endgroup$
    – admissiblecycle
    Jan 30 at 11:48

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