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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $p$ be a monic cubic polynomial such that $p(0)=1$ and such that all the zeros of $p'(x)$ are also zeros of $p(x)$ . Find $p$ . Note: monic means that the leading coefficient is 1.

Solution

Solution:

$(x+1)^3$

A root of a polynomial $p$ will be a double root if and only if it is also a root of $p'$ . Let $a$ and $b$ be the roots of $p'$ . Since $a$ and $b$ are also roots of $p$ , they are double roots of $p$ . But $p$ can have only three roots, so $a = b$ and $a$ becomes a double root of $p'$ . This makes $p'(x) = 3c(x-a)^2$ for some constant $3c$ , and thus $p(x) = c(x-a)^3 + d$ . Because $a$ is a root of $p$ and $p$ is monic, $d = 0$ and $c = 1$ . From $p(0) = 1$ we get $p(x) = (x+1)^3$ .

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