Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Philippines

Problem:
Suppose that the polynomial $P(x) = x^{3} + 4x^{2} + bx + c$ has a single root $r$ and a double root $s$ for some distinct real numbers $r$ and $s$ . Given that $P(-2s) = 324$ , what is the sum of all possible values of $|c|$ ?

Solution

Solution:
By Vieta's formula, we have $r + 2s = -4$ and writing $P(x) = (x - r)(x - s)^{2}$ , we have $324 = P(-2s) = (-2s - r)(-3s)^{2} = 36s^{2}$ . Thus, $s^{2} = 9$ and $s \in \{-3, 3\}$ .

We next observe that $|c| = |P(0)| = |r| s^{2} = |-4 - 2s| s^{2}$ . Hence, the sum of all possible values of $|c|$ is $9(|-4 - 6| + |-4 + 6|) = 108$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.