# Polynomial cannot have all roots real?

Let $$P \in \mathbb R[x]$$ be a degree-$$n$$ polynomial with real coefficients such that $$P(a) \neq 0$$, where $$a$$ is real. If $$P'(a) = P ''(a) = 0$$ then prove that $$P$$ cannot have all roots real.

Can someone suggest a possible solution using Rolle's Theorem? All I could gather was that $$P'(x) = 0$$ has a repeated root by Rolle's Theorem. But I am stuck after this.

• If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^{(n)}$ has some real root, which is impossible, cause $P^{(n)}$ is a nonzero constant.
– xbh
Aug 16, 2018 at 7:25
• @Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)\neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^{(n)}$ would have root. Am I correct now?
– xbh
Aug 16, 2018 at 7:31
• @xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^{(n)}$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial. Aug 16, 2018 at 7:39
• @Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary.
– xbh
Aug 16, 2018 at 7:44

Assume that $P$ has degree $n$ and let $x_1,x_2,\dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=c\prod_{k=1}^n (x-x_k)$, and if $x$ is not a root of $P$ we have that $$\frac{P'(x)}{P(x)}=\sum_{k=1}^n \frac{1}{x-x_k}.$$ After taking the derivative we obtain $$\frac{P''(x)P(x)-(P'(x))^2}{(P(x))^2}=-\sum_{k=1}^n \frac{1}{(x-x_k)^2}.$$ Finally by letting $x=a$ (which is not a root) we get a contradiction: $$0=\frac{P''(a)P(a)-(P'(a))^2}{(P(a))^2}=-\sum_{k=1}^n \frac{1}{(a-x_k)^2}<0$$ where the right-hand side is negative because $a, x_1,x_2,\dots,x _n$ are all real.

• Brilliant solution!
– xbh
Aug 16, 2018 at 7:58
• How dis you get the first equation? Aug 16, 2018 at 8:39
• @Szeto It is the Logarithmic derivative: en.wikipedia.org/wiki/Logarithmic_derivative Aug 16, 2018 at 8:42
• Beautiful solution! $Szeto you may want to take a look here: math.stackexchange.com/q/2660247/515527 Aug 16, 2018 at 20:12 • @Nyssa Thanks for your support. Nov 28, 2019 at 20:01 Sketch of proof: Assume all roots of$P$are real, and let$x_1\leq x_2\leq \ldots\leq x_n$be the$n$roots (with repetition if$P$has repeated roots). What does Rolle's theorem say about the roots of$P'$? How many roots does$P'$have (counted with multiplicity)? Can$P'$have a repeated root which is not one of the$x_i$? I guess$P$is not constant, otherwise the statement is false: all the roots of the constant$a$polynomial are real, as everything holds for the elements of the emptyset. Translate the polynomial by$a$, i.e.,$Q(x):= P(x-a)$. Then the conditions can be rephrased to$Q$equivalently as follows:$Q(0)\neq 0$,$Q'(0)=Q''(0)=0$. In other words,$Q(x)= a_nx^n+ \cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where$a_0\neq 0$and$a_1=a_2=0$. Write the Viete formulas for the roots$x_i$:$\prod x_i= (-1)^na_0\neq 0, \prod x_i \cdot \sum 1/x_i=0$, and$\prod x_i \cdot \sum\limits_{i\neq j} 1/(x_ix_j)=0$. Put$y_i=1/x_i$(possible, as$0$is not a root, as the constant term is nonzero), then after simplification, you obtain$\sum y_i= \sum\limits_{i\neq j} y_iy_j=0$, but then$\sum y_i^2= 0$. So if these are real numbers, then all the$y_i\$ are zero, a contradiction.

A comment:

If at a point we had two consecutive derivatives $$0$$, then the polynomial cannot have all roots real. That is: if $$P^{(k+1)}(a) = P^{(k+2)}(a)= 0$$, but $$P^{(k)}(a)\ne 0$$, then $$P^{(k)}(x)$$ cannot have all roots real, and so can't $$P(x)$$.

We may assume $$a=0$$, (consider $$Q(x) = P(x+a)$$ instead). Then the statement is: a polynomial with a gap in its coefficients cannot have all roots real (the old Théorème des lacunes).