5. Let $P(x)$ be the polynomial of minimal degree such that $P(k)=720 k /\left(k^{2}-1\right)$ for $k \in\{2,3,4,5\}$. Find the value of $P(6)$.
Answer: 48
Solution: Let $Q(x)=\left(x^{2}-1\right) P(x)-720 x$. Then $Q(k)=0$ for $k \in\{2,3,4,5\}$ so $Q(x)=R(x)(x-$ $2)(x-3)(x-4)(x-5)$ for some polynomial $R(x)$. Observe that $Q(1)=-720$ and $Q(-1)=720$, so $R(x)$ cannot be constant. As the degree of $Q$ is as small as possible, we set $R(x)=a x+b$ for some constants $a$ and $b$ so that $Q(x) \stackrel{(\star)}{=}(a x+b)(x-2)(x-3)(x-4)(x-5)$. Letting $x=1$ on $(\star)$, we have $a+b \stackrel{(1)}{=}-30$ and letting $x=-1$ on $(\star)$, we have $-a+b=2$. From (1) and (2), we get $(a, b)=(-16,-14)$. Thus,
$$
\left(x^{2}-1\right) P(x)-720 x=Q(x)=(-16 x-14)(x-2)(x-3)(x-4)(x-5)
$$
and letting $x=6$ yields $35 P(6)-720(6)=-110(24)=-2640$. Hence, $P(6)=48$.