$4 \cdot 9$ Given that a quadratic trinomial achieves an extremum of 25 when $x=\frac{1}{2}$; the sum of the cubes of the roots of this quadratic trinomial is 19. Try to find this quadratic trinomial.
[Solution] From the first given condition, we know that the leading coefficient of this quadratic trinomial is not equal to 0, and it can be written as
$$a\left(x-\frac{1}{2}\right)^{2}+25$$

which is $a x^{2}-a x+\frac{1}{4} a+25$.
Thus, by Vieta's formulas, we have
$$x_{1}+x_{2}=1, x_{1} x_{2}=\frac{1}{4}+\frac{25}{a} \text {. }$$

Therefore,
$$\begin{aligned}
x_{1}^{3}+x_{2}^{3} & =\left(x_{1}+x_{2}\right)\left[\left(x_{1}+\right.\right. \\
& =1-3\left(\frac{1}{4}+\frac{25}{a}\right) \\
& =\frac{1}{4}-\frac{75}{a} .
\end{aligned}$$

From the second given condition, we get $\frac{1}{4}-\frac{75}{a}=19$,
so $a=-4$.
Substituting $a=-4$ into (1), we get $-4 x^{2}+4 x+24$.
Upon verification, the quadratic trinomial $-4 x^{2}+4 x+24$ is the desired result.