5. If the real number $k$ makes the equation about $x$
$$
\left(x^{2}-1\right)\left(k x^{2}-6 x-8\right)=0
$$

have exactly three distinct real roots, then $k$ is called a "good number". The number of such good numbers $k$ is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4
5. D.

From the condition, we know that $x= \pm 1$ are roots of the equation.
If $x=1$ is a root of the equation
$$
k x^{2}-6 x-8=0
$$

we get $k=14$, and the other root is $-\frac{4}{7}$, which satisfies the condition.
If $x=-1$ is a root of equation (1), we get $k=2$, and the other root is -4, which satisfies the condition.
If $x= \pm 1$ are not roots of equation (1):
When $k=0$, the root of equation (1) is $-\frac{4}{3}$, which satisfies the condition;
When $k \neq 0$, the discriminant of equation (1) is
$$
\Delta=(-6)^{2}+32 k=0 \Rightarrow k=-\frac{9}{8} \text {. }
$$

In this case, equation (1) has two equal real roots $-\frac{8}{3}$, which satisfies the condition.
Therefore, there are 4 good numbers $k$.