4. Let $x_{1}, x_{2}$ be the two roots of the quadratic equation $x^{2}+x-3=0$. Then, the value of $x_{1}^{3}-4 x_{2}^{2}+19$ is ().
(A) -4
(B) 8
(C) 6
(D) 0
4. (D).
$\because x_{1}+x_{2}$ are the roots of the quadratic equation $x^{2}+x-3=0$,
$$
\therefore x_{1}^{2}+x_{1}-3=0, x_{2}^{2}+x_{2}-3=0 \text {, }
$$

i.e., $x_{1}^{2}=3-x_{1}, x_{2}^{2}=3-x_{2}$.
By the relationship between roots and coefficients, we know $x_{1}+x_{2}=-1$, thus we have
$$
\begin{array}{l}
x_{1}^{3}-4 x_{2}^{2}+19=x_{1}\left(3-x_{1}\right)-4\left(3-x_{2}\right)+19 \\
=3 x_{1}-x_{1}^{2}+4 x_{2}+7 \\
=3 x_{1}-\left(3-x_{1}\right)+4 x_{2}+7 \\
=4\left(x_{1}+x_{2}\right)+4=4(-1)+4=0 .
\end{array}
$$