Number of rational roots of the equation x^{2 | 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,for a polynomial with integer coefficients $a_n x^n + \dots + a_0 = 0$,any rational root $\frac{p}{q}$ must satisfy $p \mid a_0$ and $q \mid a_n$.Here,$a_n = 1$ and $a_0 = 1$.Thus,$p \mid 1$ and $q \mid 1$,which implies $p, q \in \{-1, 1\}$.Therefore,the only possible rational roots are $x = 1$ or $x = -1$.Let $f(x) = x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1$.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.Thus,the number of rational roots is $0$.

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

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