4. [4] Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
Answer: $\frac{89}{11}$
Solution: Observe that 7 is not a root of $P$. If $r_{1}, r_{2}, r_{3}, r_{4}$ are the roots of $P$, then $\frac{P^{\prime}(7)}{P(7)}=$ $\sum_{i} \frac{1}{7-r_{i}}=0$. Thus $r_{4}=7-\left(\sum_{i \neq 4} \frac{1}{7-r_{i}}\right)^{-1}=7+\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)^{-1}=7+12 / 11=89 / 11$.