Let $\eta_i,0\leq i\leq 4$ be 4 variables and $\eta_0 x^2+4\eta_1 x^3y+6\eta_2 x^2y^2+4\eta_3 xy^3+\eta_4 y^4$ be the associated quadratic form. Consider $U=x^2,V=2xy, W=y^2$. Then the quadratic form transforms into $$\eta_0 U^2+2\eta_2UV+\eta_2(V^2+2UW)+2\eta_3VW+\eta_4W^2=0\quad (1)$$ and there is an additional relation $$4UW-V^2=0\quad (2)$$ Denote by $Q_1$ and $Q_2$ the associated quadratic forms of $(1)$ and $(2)$.
Consider $\det(Q_1+\lambda Q_2)=0$ for solving $\lambda$ (i.e. $(1)+\lambda(2)=0$ is associated quadratic form). Then plugging $\lambda$ in $(U,V,W)^T(Q_1+\lambda Q_2)(U,V,W)$, one will see the polynomial splits into linear factors.
$\textbf{Q:}$ Why does the polynomial splits into linear factors? I could see there is a coordinate transformation s.t. $Q_1+\lambda Q_2$ is effectively (2 by 2 matrix) $\oplus\ 0$ by $\det(Q_1+\lambda Q_2)=0$. That 2 by 2 is quadratic as well. Hence it reduces to quadratic of 2 variables which splits over complex number. This seems very cumbersome.
Ref. An Introduction to Invariants and Moduli, Mukai, Remark 1.26 of Chapter 1, Sec. 3(b), pg 28.