Question 27. Given the quadratic equation in $\mathbf{x}$: $x^{2}+2 p \mathbf{x}-6 \mathbf{x}+2 p^{2}-5 p-3=0$ has real roots, find the maximum and minimum values of the product of the roots.
Let the product of the two roots be $y$, by Vieta's formulas,
$$
\mathrm{y}=2 \mathrm{p}^{2}-5 \mathrm{p}-3=2\left(\mathrm{p}-\frac{5}{4}\right)^{2}-\frac{49}{8} \text {. }
$$

Given that the original equation has real roots,
$$
\begin{array}{l} 
\therefore \quad(2 p-6)^{2}-4\left(2 p^{2}-5 p-3\right) \geqslant 0, \\
p^{2}+p-12 \leqslant 0 . \\
-4 \leqslant p \leqslant 3 .
\end{array}
$$
$\because$ The quadratic function of $p$, $y$, is a parabola symmetric about the line $p=\frac{5}{4}$. The distance between $(-4,0)$ and $\left(\frac{5}{4}, 0\right)$ is greater than the distance between $\left(\frac{5}{4}, 0\right)$ and $(3,0)$.
$\therefore$ When $p=-4$, the product of the roots of the original equation $y_{\text{max}}=49$. When $\mathrm{P}=\frac{5}{4}$, the product of the roots of the original equation $y_{\text{min}}=-\frac{49}{8}$.