2. The quartic (4th-degree) polynomial $P(x)$ satisfies $P(1)=0$ and attains its maximum value of 3 at both $x=2$ and $x=3$. Compute $P(5)$.
2. Consider the polynomial $Q(x)=P(x)-3$. Then $Q(x)$ has zeros at and maximum value 0 at $x=2,3$. These conditions imply that $Q(x)$ has the form
$$
Q(x)=A(x-2)^{2}(x-3)^{2} \text {. }
$$

That is, its graph looks like because the values of $Q(x)$ should grow larger and larger through negative values as the variable $x$ goes to larger and larger values of both signs and the fact that the number of turning points should not exceed $4-1=3$ but should be more than 2 (given by the maximum points). Thus, $Q(1)=-3$ imply $a=-\frac{3}{4}$. Finally, $Q(5)=P(5)-3=-27$ and $P(5)=-24$.