The number of rational roots of the equation | vedclass

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Question

(A) Let $x = \frac{p}{q}$ be a rational root in simplest form,where $p, q \in \mathbb{Z}$,$q > 0$,and $\gcd(p, q) = 1$. According to the Rational Root

Vedclass · Accepted Answer

$0$

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(A) Let $x = \frac{p}{q}$ be a rational root in simplest form,where $p, q \in \mathbb{Z}$,$q > 0$,and $\gcd(p, q) = 1$. According to the Rational Root Theorem,$p$ must be a divisor of the constant term $1$,and $q$ must be a divisor of the leading coefficient $1$. Thus,$p, q \in \{-1, 1\}$,which implies $x \in \{-1, 1\}$. Now,test these values in the equation $f(x) = x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$: For $x = 1$: $f(1) = 1^{2016} - 1^{2015} + 1^{1008} + 1^{1003} + 1 = 1 - 1 + 1 + 1 + 1 = 3 
eq 0$. For $x = -1$: $f(-1) = (-1)^{2016} - (-1)^{2015} + (-1)^{1008} + (-1)^{1003} + 1 = 1 - (-1) + 1 - 1 + 1 = 1 + 1 + 1 - 1 + 1 = 3 
eq 0$. Since neither $1$ nor $-1$ is a root,the equation has no rational roots. Therefore,the number of rational roots is $0$.

The number of rational roots of the equation $x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$ is equal to

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