4.9 Given that a quadratic trinomial attains 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
$$
a\left(x-\frac{1}{2}\right)^{2}+25 \text {, }
$$

which is $\square$
$$
a x^{2}-a x+\frac{1}{4} a+25 .
$$

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 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.