Given a polynomial that is not constant, of course, it doesn’t contain a constant interval. But how can we prove it?

Since the polynomial is differentiable over R, I came up with a solution that uses Lagrange mean value theorem n times and reduces the nth derivative to a constant. Since the leading term is not zero, there can not be a zero in the nth derivative, and that contradicts the mean value theorem. Therefore, the polynomial must not contain a constant interval.

However, I do realize that this solution is a bit complicated, so is there a simpler solution(possibly elementary) that can prove this?

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    $\begingroup$ If the polynomial is constant in an interval, then by subtracting the constant from the polynomial, the new polynomial is zero in that interval. i.e. the polynomial has an infinite number of roots. $\endgroup$
    – peterwhy
    Aug 11, 2021 at 18:55

3 Answers 3


Suppose $P(x) = c$ for all $x \in [a, b]$ with $a < b$ and $P$ is nonconstant. Then let $n > 0$ be the degree of the polynomial. Then $Q(x) = P(x) - c$ is also of degree $n$. But $Q$ has roots $a + \frac{i}{n} (b - a)$ for $0 \leq i \leq n$, which is $n + 1$ roots. And a polynomial of degree $n$ with coefficients in a field can have at most $n$ roots. Contradiction.


A polynomial of degree $n \gt 0$ has at most $n$ roots. A constant polynomial is of degree $0$.

A non constant polynomial $p$… can’t be constant. Therefore it is of degree $n\gt 0$. If it would be constant and equal to $a$ on an interval of strictly positive lenght, $p-a$ would have an infinite number of roots. A contradiction.


Suppose $p$ is a polynomial of degree $n$, i.e. $$ p(x) = \sum\limits_{i=0}^n a_ix^i $$

and $p(x) = c$ for $x\in (a,b)$. Choose $n+1$ distinct points from $(a,b)$, $x_0,\dots , x_n$. The system of equations $p(x_i) = c$ for the $a_i$ is clearly solved by $a_0 = c$ and $a_i = 0$ otherwise. However, since the matrix of coefficients (the Vandermonde matrix) is invertible whenever the $x_i$ are distinct*, the solution is unique. Thus, $p(x) \equiv c$.


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