College Algebra, Chapter 4, 4.4, Section 4.4, Problem 86

Prove that the polynomial $P(x) = x^{50}-5x^{25}+x^2-1$ does not have any rational zeros.
Since the leading coefficient is $1$, any rational zero must be a divisor of the constant term $-1$. So the possible rational zeros are $\pm 1$. We test each of these possibilities


$
\begin{equation}
\begin{aligned}
P(1) &= (1)^{50}- 5(1)^{25} + (1)^2 - 1\\
\\
P(1) &= -4\\
\\
P(1) &= (-1)^{50} -5 (-1)^{25}+ (-1)^2 - 1\\
\\
P(1) &= 6
\end{aligned}
\end{equation}
$


By lower and upper bounds theorem, $-1$ is the lower bound $1$ is the upper bound for the zeros. Since neither $-1$ nor $1$ is a zerom all the real zero lie between these numbers.