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I'm trying to show that the following function has no zeros $$ 60x^4-44x^3-25x^2-44x+60=0. $$ I already tried using Eisenstein's criterium, but since the first and the last coefficient are both $60$, there is no prime which holds.

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In this answer I showed that the polynomial

$$ p(x) = x^4 + ax^3 + bx^2 + ax + 1 $$

has no real zeros if $|a| \leq 4$ and $b > 2 |a| - 2$.

We have

$$ 60x^4-44x^3-25x^2-44x+60 = 60 \left(x^4 - \frac{44}{60}x^3 - \frac{25}{60}x^2 - \frac{44}{60}x+1\right), $$

so in this case $a = -44/60$ and $b = -25/60$. Indeed, $|a| \leq 4$ and

$$ b = -\frac{25}{60} > -\frac{32}{60} = 2|a|-2, $$

so the given polynomial has no real zeros.

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Following the hint in the comments, we divide by $x^2$ and rearrange to get $$60\left(x^2+\frac{1}{x^2}\right)-44\left(x+\frac{1}{x}\right)-25=0$$ Set $y=x+\frac{1}{x}$ and note that $y^2=x^2+\frac{1}{x^2}+2$ to get $$60(y^2-2)-44y-25=0$$ which rearranges to the quadratic $$60y^2-44y-145=0$$ This has roots $y\approx -1.23, 1.96$. [Alternate answer: $g(y)=60y^2-44y-145$ is an upward-pointing parabola, and $g(2)>0$ and $g(-2)>0$, so any roots are in $(-2,2)$.]

However $|x+\frac{1}{x}|\ge 2$ for all real $x$, so there are no real solutions.

Cute proof of the last fact: for $x>0$, the arithmetic-geometric mean inequality gives $x+\frac{1}{x}\ge 2\sqrt{x\cdot \frac{1}{x}}=2$.

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