Four. (Full marks 20 points) Let the function $f(x)=a x^{2}+8 x+3(a<0)$. For a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$.

Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$. Prove your conclusion.
Given, $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$, so,
$$
\max _{x \in R} f(x)=3-\frac{16}{a} \text {. }
$$

We discuss in two cases:
(i) $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a} \text {. }
$$

Thus, $l(a)$ is the larger root of the quadratic equation $a x^{2}+8 x+3=-5$,
$$
\begin{aligned}
l(a) & =\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{4}{\sqrt{4-2 a}-2} \\
& \leqslant \frac{4}{\sqrt{2}-2}=\frac{\sqrt{5}+1}{2},
\end{aligned}
$$

Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, $l(a)$ attains its maximum value $\frac{\sqrt{5}+1}{2}$ if and only if $a=-8$.