4. The root situation of the equation $x^{3}+x^{2}+x-2=0$ is ( ).
(A) There is one positive real root
(B) There are two positive real roots
(C) There is one negative real root
(D) There are two negative real roots
4. A.

Obviously, $x \neq 0$.
Thus, $\frac{x^{3}+x^{2}+x-2}{x}=0$, which means $x^{2}+x+1=\frac{2}{x}$. Let $y_{1}=x^{2}+x+1$, $y_{2}=\frac{2}{x}$. As shown in Figure 6, plot the graphs of the two functions. From the graph, we can see that the two function graphs intersect at only one point and it is in the first quadrant. Therefore, the original equation has one positive real root.