Example 7 When $m=$ $\qquad$, the polynomial $6 x^{2}+$ $m x y-4 y^{2}-x+17 y-15$ can be factored into the product of two binary linear trinomials in $x$ and $y$.
(1990, Hope Cup Junior High School Mathematics Invitational Competition)
Solution: $\because 6 x^{2}+m x y-4 y^{2}-x+17 y-15$
$$
\begin{array}{l}
=6 x^{2}+(m y-1) x-\left(4 y^{2}-17 y+15\right), \\
\therefore \Delta=(m y-1)^{2}+24\left(4 y^{2}-17 y+15\right) \\
=\left(m^{2}+96\right) y^{2}-2(m+204) y \\
+361. \\
\end{array}
$$

For the polynomial $6 x^{2}+m x y-4 y^{2}-x+17 y-15$ to be factorable into the product of two linear expressions in $x$ and $y$, its discriminant must be a perfect square, i.e., the discriminant of $\left(m^{2}+96\right) y^{2}-2(m+204) y+361$ must be 0, which means $[-2(m+204)]^{2}-4 \times 361 \cdot\left(m^{2}+96\right)=0$. Solving this, we get $m=5$ or $m=-\frac{58}{15}$.

Therefore, when $m=5$ or $m=-\frac{58}{15}$, the polynomial $6 x^{2}+m x y-4 y^{2}-x+17 y-15$ can be factored into the product of two linear expressions in $x$ and $y$.

Note: The discriminant $\Delta$ can be used twice to solve the real factorization problem of a real-coefficient binary quadratic polynomial of the form $a x^{2}+b x y+c y^{2}+d x+e y+f(a \neq 0)$.