# Proof verification - A polynomial $P(x)$ has only real roots $\implies$ $P'(x)$ also has only real roots

Here's a problem that I've solved but I'm not very confident on my solution. Please check it there's any gap in my arguments. Also, is there a way to come up with a shorter proof ? Thank you.

The Problem : If the polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0,~a_n \neq 0,$ $(a_j \in \mathbb{R} ~\text{ for }~0 \leq j \leq n)$ has only real roots, then it's derivative $P'(x)$ has only real roots.

My Solution : We know that, a polynomial of degree $n$ has exactly $n$ roots. Let the roots be denoted by $\alpha_1, \alpha_2,\ldots,\alpha_n$.

Case-I : Let all the roots be distinct. WLOG let $\alpha_1<\alpha_2<\ldots<\alpha_n$. Then, by Rolle's theorem $\forall~ 0 \leq j \leq n, \exists~ c_j \in (\alpha_j,\alpha_{j+1})$ such that $P'(c_j)=0$. Note that we're free to use Rolles's theorem as $P$ is differentiable $($and hence also continuous$)$ throughout $\mathbb{R}$. Hence we have $n-1$ real roots of $P'(x)$ namely $c_1,c_2,\ldots,c_{n-1}$. Since $P'(x)$ is a polynomial of degree $n-1,$ and hence has exactly $($though at most suffices, in this case$) ~n-1$ roots. Hence all roots of $P'(x)$ are real. So this case was quite trivial (in a relative sense).

Case-II : All the roots are NOT distinct. Suppose we have $m~(<n)$ distinct roots $\beta_1, \beta_2,\ldots,\beta_m$ with respective multiplicities $k_1,k_2,\ldots,k_m$. Clearly, $\sum_{j=1}^m k_j=n$.

Claim : If $\beta$ is a root of $P(x)$ with multiplicity $k~(>1),$ then $\beta$ is a root of $P'(x)$ with multiplicity $k-1$.

Proof of the claim : $\beta$ is a root of $P(x)$ with multiplicity $k~(>1)$ $\implies$ $P(x)=(x-\beta)^k Q(x)$ where $Q(x),$ is a polynomial of degree $n-k$ and $Q(\beta) \neq 0$. Then $P'(x)=k(x-\beta)^{k-1}Q(x)+(x-\beta)^kQ'(x)=(x-\beta)^{k-1}\{kQ(x)+(x-\beta)Q'(x)\}$

Hence $\beta$ is a root of $P'(x)$ with $k-1$ multiplicity $($Since $kQ(\beta)+(\beta-\beta)Q'(\beta)=kQ(\beta) \neq 0)$. End of Proof of the claim.

Using the claim, for all $0 \leq j \leq m, \beta_j$ is a root of $P'(x)$ with multiplicity $k_j-1$.

Like in Case-I, we use Rolle's theorem again to come up with $c_j \in (\beta_j,\beta_{j+1})$ such that $P(c_j)=0, ~1 \leq j \leq m-1$.

So in total, as roots of $P'(x),$ we have $c_1,c_2,\ldots,c_{m-1}$ each with multiplicity $1,$ $($hence in total $m-1$ roots$),$ and $\beta_1,\beta_2,\ldots,\beta_m$ with respective multiplicities $k_1-1,k_2-1,\ldots,k_m-1$ $($which adds up to $\sum_{j=1}^m(k_j-1)=n-m$ roots$).$ Hence we have $m-1+n-m=n-1$ real roots of $P'(x)$. Since $P'(x)$ is a polynomial of degree $n-1,$ it has exactly $n-1$ roots. Hence all the roots of $P'(x)$ is real. $\blacksquare$

• A shorter proof is a direct consequence of Gauss-Lucas theorem – Sahiba Arora Jul 2 '17 at 17:31
• @SahibaArora I have not done up to that yet but thanks anyways! – Dragon Jul 2 '17 at 17:34
• Your proof is simple, easy to understand and correct! And it is perhaps the most natural way to attack the problem. There might be a shorter proof, but it might not have the qualities your proof already possesses. +1 – Paramanand Singh Jul 3 '17 at 5:30