$2 \cdot 24$ Let $F=\frac{6 x^{2}+16 x+3 m}{6}$ be the square of a linear expression in $x$, then there must be a value of $m$ between
(A) 3 and 4.
(B) 4 and 5.
(C) 5 and 6.
(D) -4 and -3.
(E) -6 and -5.
(20th American High School Mathematics Examination, 1969)
[Solution] The given expression for $F$ can be rewritten as
$$
F=x^{2}+\frac{8}{3} x+\frac{m}{2}=\left(x+\frac{4}{3}\right)^{2}+\left(\frac{m}{2}-\frac{16}{9}\right) \text {. }
$$

When $m=\frac{32}{9}$, $F=\left(x+\frac{4}{3}\right)^{2}$.
That is, $\square$
$$
3<m<4 \text {. }
$$

Therefore, the correct choice is $(A)$.