According to the book of Harris, Algebraic geometry, A first course, page $9$, the twisted cubic is defined to be the image $C$ of the map $ v : \mathbb{P}^1 \to \mathbb{P}^3 $ given by $$ v : [X_0 , X_1 ] \to [ X_{0}^{3} , X_{0}^{2} X_{1} , X_{0} X_{1}^{2} , X_{1}^{3} ].$$
The author of this book says then, that $C$ lies on the $3$ quadric surfaces $ Q_0 $ , $ Q_1 $ and $ Q_2 $ given as the zero locus of the polynomials $ F_0 (Z) = Z_0 Z_2 - Z_{1}^2 $ , $ F_1 (Z) = Z_0 Z_3 - Z_{1} Z_2 $ and $ F_2 (Z) = Z_0 Z_3 - Z_{2}^2 $.
Could you explain to me why do we have this? How do we obtain those polynomials in that form?
Thanks in advance for your help.