TITLE: How to prove that $\frac{d^n}{dx^n}(x^2-1)^n=0$ has $n$ real roots?
QUESTION [11 upvotes]: How do I prove that $$\frac{d^n}{dx^n}(x^2-1)^n=0$$ has $n$ real roots?

REPLY [1 votes]: Let $P(x)$ be a polynomial of degree $n$. Suppose $P(x)$ has a zero of degree $m$ at $x=a$. This means that
$$
P(x)=(x-a)^mQ(x)
$$
where $Q(x)$ is a polynomial of degree $n-m$ and $Q(a)\ne0$. Thus,
$$
\begin{align}
P'(x)
&=(x-a)^{m-1}\left[mQ(x)+(x-a)Q'(x)\right]\\
&=(x-a)^{m-1}R(x)
\end{align}
$$
where $R(x)$ is degree $n-m$ and $R(a)=mQ(a)\ne0$.
That is, each zero of $P(x)$ is reduced by one degree. However, Rolle's Theorem insures that between two adjacent zeros of $P(x)$, $P'(x)$ has a zero.
Let $k$ be the number of zeros of $P(x)$ not counting multiplicities. Differentiation will reduce the degree of each of the $k$ zeros by one, and will add $k-1$ zeros between the $k$ zeros by Rolle's Theorem. Thus, the total number of zeros of $P'(x)$, counting multiplicities, is at least the number of zeros of $P(x)$, counting multiplicities, minus one. New zeros might be introduced to $P'(x)$.
The number of zeros not counting multiplicities of $P'(x)$ is at least $k-1$. In addition, each multiple zero of $P(x)$ persists. Thus, $P'(x)$ has at least $k+m-1$ roots not counting multiplicities where $P(x)$ has $k$ zeros, not counting multiplicities including $m$ multiple zeros.
Counting multiplicities, $(x^2-1)^n$ has $2$ zeros of degree $n$. Therefore, these two roots persist until the $n^{\text{th}}$ derivtive. After $n-1$ derivatives, there should be at least $n+1$ roots with the multiple roots gone. After the $n^{\text{th}}$ derivative, there should be at least $n$ roots, not counting multiplicity. As a degree $n$ polynomial, it has at most $n$ roots. Therefore, it must have exactly $n$ roots.
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