19. The number of rational values that make $f(x)=-2 x^{5}+4 x^{3}-2 x^{2}+5$ equal to 0 is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) $\neq$
19. (E).

Let $f(x)=0$ have a rational root $x_{1}=\frac{p}{q}$. Where $p, q \in \mathbb{Z}$. $(p, q)=1, q \neq 0$. Then
$$
f\left(x_{1}\right)=-2\left(\frac{p}{q}\right)^{5}+4\left(\frac{p}{q}\right)^{3}-2\left(\frac{p}{q}\right)^{2}+5=0 .
$$

That is, $2 p^{5}=\left(4 p^{3}-2 p^{2} q+5 q^{3}\right) q^{2}$.
Thus, $q^{2} \mid 2 p^{5}$. But $(p, q)=1$. Therefore, $q^{2} \mid 2$. Hence, $q=1$.
Therefore, any rational root $x_{1}$ of $f(x)=0$ is an integer, i.e., $x_{1} \in \mathbb{Z}$.
But $f\left(x_{1}\right)=-2 x_{1}^{5}+4 x_{1}^{3}-2 x_{1}^{2}+5 \equiv 1 \pmod{2}$. This contradicts $f(x)=0$. Hence, $f(x)=0$ has no rational roots.