HMMT-Feb 2010 calc, Problem 3

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

Solution: A root of a polynomial $p$ will be a double root if and only if it is also a root of $p^{\prime}$. Let $a$ and $b$ be the roots of $p^{\prime}$. 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 $p^{\prime}$. This makes $p^{\prime}(x)=3 c(x-a)^{2}$ for some constant $3 c$, 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 $p(x)=(x+1)^{3}$. $\fbox{(x+1)^3}$.

Answer: (x+1)^3