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Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Let $P$ be a fourth degree polynomial, with derivative $P'$ , such that $P(1) = P(3) = P(5) = P'(7) = 0$ . Find the real number $x \neq 1, 3, 5$ such that $P(x) = 0$ .

Solution

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
$\frac{P'(7)}{P(7)} = \sum_{i} \frac{1}{7 - r_i} = 0.$
Thus
$r_4 = 7 - \left( \sum_{i \neq 4} \frac{1}{7 - r_i} \right)^{-1} = 7 + \left( \frac{1}{6} + \frac{1}{4} + \frac{1}{2} \right)^{-1} = 7 + \frac{12}{11} = \frac{89}{11}.$

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