# If the polynomial equation

Question:

If the polynomial equation

$a_{0} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}=0$

n positive integer, has two different real roots α and β, then between α and β, the equation

$n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ has

(a) exactly one root
(b) almost one root
(c) at least one root
(d) no root

Solution:

(c) at least one root

We observe that, $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ is the derivative of the polynomial $a_{n} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}=0$

Polynomial function is continuous every where in R and consequently derivative in R

Therefore, $a_{n} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$ is continuous on $[\alpha, \beta]$ and derivative on $(\alpha, \beta)$. Hence, it satisfies the both the conditions of Rolle's theorem.

By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function $f(x)$, there exists at least one root of its derivative.

Hence, the equation $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ will have at least one root between $\alpha$ and $\beta$.