Let

$$
K=\frac{\left(m^{2}+2 m-24\right) x^{2}-6(m+6) x+\left(m^{2}+10 m+24\right)}{\left(m^{2}-9 m+18\right) x^{2}-8(m-6) x+\left(m^{2}-3 m-18\right)}
$$

Determine all values of $m$ for which a) $K>0$ for all $x$; b) $K<0$ for all $x$.
For a fixed $m$, condition (a) or (b) can only be satisfied for all $x$ if $K$ is defined for all $x$ and its value is never zero. This follows if neither the numerator nor the denominator of $K$ is zero, meaning the numerator and the denominator - and thus $K$ itself - are of constant sign. The numerator and denominator are of constant sign precisely when the discriminants of the two quadratic polynomials in the numerator and denominator are negative:

\[
\begin{gathered}
6^{2}(m+6)^{2}-4\left(m^{2}+10 m+24\right)\left(m^{2}+2 m-24\right)=(m+6)^{2}\left(100-4 m^{2}\right)5
\end{gathered}
\]

We know, therefore, that (2) must be satisfied for either (a) or (b) to hold, and we also know that if (2) is satisfied, then the value of $K$ is either positive for all $x$ or negative for all $x$. Thus, to determine for which values of $m$ that satisfy (2), $K$ will be positive or negative, we can check the sign of $K$ at any arbitrary $x$. Let, for example, $x=0$, then

\[
K(0)=\frac{m^{2}+10 m+24}{m^{2}-3 m-18}=\frac{(m+4)(m+6)}{(m-6)(m+3)}
\]

It can be seen that the numerator is positive if $-4 < m < -3$ or $m > 6$; and the denominator is positive if $m < -3$ or $m > 6$. Therefore:

(a) $K$ is exactly positive for all $x$ if $-4 < m < -3$ or $m > 6$;

(b) $K$ is exactly negative for all $x$ if $5 < |m| < 6$.